If we have a point charge $q$ then at any point around $q$ we have an electric field due to our source charge $q$. We have a certain value of voltage or potential at this point, which is nothing but the work done by $q$ in moving a charge of $+1C$ from my reference point to infinity.   Is my understanding correct?
123442
The english wikipedia page defines the electic potential in the following way:     The electric potential at a point r in a static electric field E is   given by the line integral      $V_\mathbf{E} = - \int_C \mathbf{E} \cdot \mathrm{d}  \boldsymbol{\ell}$      where $C$ is an arbitrary path connecting the point with zero   potential to $r$.   So if we take infinity to be a point with zero potential, the path we take is from the infinity to that point.  However the other definition I know states that the electric potential at a given point is equal to the work that needs to be done to move charge $q$ from that point to infinity, divided by $q$.  It seems to me that the definitions are not equivalent (both give the same result, but with opposite sign). Could anyone clarify this for me please?
89382
In the  (semi-classical, but should still apply here I think), the conducting electrons are in a constant electric field, and, in between collisions with the lattice ions (that happen, on average, in a time $\tau$ since their last collision) in which their velocities are reset to an average of 0, they are accelerated and gain a momentum  $$p = qE\tau (=mv)$$, where $q$ is its charge and $E$ is the electric field from the applied voltage. This obviously means in between the collisions, the electron accelerates $a$, from $v_f = v_i + a\tau$.  Accelerating charges give off . So does a DC carrying wire give off radiation one can measure?  I suspect it does, but it's of an insanely small magnitude. A quick order of magnitude calculation with the following values gives me:  $$q \sim 10^{-19} \,{\rm C}, \\ E\sim 1\,{\rm V}/1\,{\rm cm}=100\,{\rm V/meter},\\ m\sim10^{-30}\,{\rm kg} \\ \Rightarrow a = qE/m \sim 10^{13}{\rm m/s}^2 \\ \Rightarrow P \sim q^2a^2/\epsilon_0c^3 \sim 10^{-28}\,{\rm W}$$  Which is obviously pretty small. I'm not sure what's the smallest power that can be measured, but I'm guessing by increasing the applied voltage one could increase the radiated power. But is the idea even fundamentally sound?
114044
I have come to know the electrostatic potential energy in vacuum is given by $${\frac{1}{2}} \epsilon_0\int d^3x {E^2} $$ and this energy  is due to the mutual electrostatic coulomb potential energy. So only a single point charge in space also contributes to some electrostatic potential energy according to $${\frac{1}{2}} \epsilon_0 \int d^3x {E^2} $$  as due to a single point charge also we have an electric field at every point in space. Now I am not getting it physically, why should there be any Coulomb energy in this case as there is no other charge to provide the electrostatic coulomb potential energy?
33954
Suppose you place a point charge $+Q$ at the corner of an imaginary cube.  Since electric field lines are radial, there is no flux through the three adjacent (adjacent to the charge) sides of the cube. However there is some amount of flux passing through the other three sides of the cube (flowing out of the cube).  We can estimate that the flux through these three surfaces combined is equal to $Q/(8\epsilon)$. As, if you consider $7$ other cubes having the charge at the corner, each of them would have equal flux flowing out by symmetry and since the total flux through the $8$ cubes is $Q/\epsilon$, each cube would have a flux of $Q/(8\epsilon)$.  Now apply Gauss' law to the cube, and we find that the cube encloses a charge of $Q/8$.  This means that 1/8th of the charge belongs to this cube.   But the charge we placed was a point charge with no dimensions. It cannot be split into parts.  What is wrong?
124240
I will try to limit the question in the case of the electric fields, but is something that applies also to the magnetic ones. There are two ways to express the energy in a capacitor:    By Voltage : $U = 1/2 CV^2 $ And by Field : $U = 1/2 \varepsilon E^2Ad$, With Energy Density: $u = 1/2 \varepsilon E^2 $    Unless i understood everything wrong and these two are NOT the same quantity, i have the following question. When we have two charges placed at points A and B, then in order to calculate the energy of the system, we will take the first charge, place it at point A WITHOUT doing any work, and then we will calculate the work needed to place the second charge at point B.  The weird thing to me here, is that while we have placed the first charge, without generating any work the system will still have the energy held in the field of the charge!   There is obviously something that i miss, but what?
5956
If the charge $q_1$ has to repel the charge $q_2$, the electric field has to go inside the conductor which contradicts the fact that electric field inside conductors is zero. Then why do the charges distribute themselves in a particular manner? Why can't they be distributed over a small place on the conductor?
8080
When we're dealing with problems in electrostatics (especially when we use Gauss' law) we often refer to the density of electric , which is inversely proportional to the radius in the case of a single point charge (all field lines are directed radially).   My question may sound dumb, despite the fact that this concept is quite intuitive, but if you think about it there is actually an infinite amount of field lines which we can draw everywhere (one through every point), so speaking of an area where field lines are more "dense" or "sparse" doesn't make much sense to me.  With this in mind, why can we still use such a concept? Why does it really work?
78037
I have, for instance, a problem with a spherically symmetric charge distribution. I deduce here, in order to solve the problem easily, that the corresponding electric field must be symmetric. How is this type of deduction rigorously justified?
96843
I am an electronics beginner and I created a small little circuit, just a battery a resistor and a led, then to ground. This is the circuit:         &ndash; Schematic created using   Now I have a few questions about it. I know voltage is the different of electrical charge between two points, so I measured that difference:  Point A to B: 3.27V Point B to C: 1.82V   First question: Does that mean that the actual voltage of: Point A: 5V Point B: 1.73V Point C: 0V?   So I am not sure if my calculations of voltage are correct. I also have this other questions:   Does the resistor have to be before the component I am trying to protect, like an LED. I have seen people place it after, but wouldn't that give a full 5V to the led or component? Ground has a neutral charge right? so why do electrons want to travel so badly to a point that has no electric charge?   Thank you
79468
I have a questions about the electric field lines.   Well in the basic learning, we know that:   The electric field lines extend away from a positive charge They move forward a negative charge   Let's take parallel plates, which make a uniform electric field.If we take the basic learning, which I mentioned, in accounts, it's very easy to understand that this is the direction of a positive charge object if we put the object between the plate (move away the positive charged plate and toward negatively charged plate).  The problem is if we put a negative charged object between. Isn't the electric field reversed ?
45877
If a capacitor is charged by a source with constant voltage, $V$, the energy, $W$, stored by a capacitor of capacitance $C$ is given by:  $$ W = \frac{1}{2} QV = \frac{1}{2} V^{2}C = \frac{Q^{2}}{2C} $$  where $Q$ is the charge stored on the plates.  If the capacitor is now disconnected, and its capacitance is changed by increasing the distance between the plates, its energy must increase since work is being done by moving charge under the influence of an electrical field. Furthermore, since the plates are not connected to anything, it can be assumed that charge is conserved.  The method for calculating the change in energy that I have seen in textbooks and on the internet is to calculate $Q$ for the pre-separated case, calculate $C$ for both cases, then use:  $$ W = \frac{Q^{2}}{2C} $$  to give $$ \delta W = \frac{Q^{2}}{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$  where $C_{2}$ is capacitance post-separation, $C_{1}$ is capacitance pre-separation and $\delta W$ is the change in energy.  This seems naive to me. That formula was for the process of charging a capacitor from a constant voltage with the $\frac{1}{2}$ arising from the potential difference that charge is raised by decreasing as more charge is deposited.  With constant $Q$, we can say that:  $$ dW = QdV $$  and integrate to give:  $$ \delta W = Q(V_{2} - V_{1}) $$  where $V_{2}$ is the post-separation voltage and $V_{1}$ is the pre-separation voltage. Substituting $Q = CV$ then gives:  $$ \delta W = Q^{2} \left( \frac{1}{C_{2}} - \frac{1}{C_{1}} \right) $$  i.e. double the textbook answer.  Am I missing something?
29129
Since the electric field inside a conductor is zero that means the potential is constant inside a conductor, which means the "inside" of a conductor is an equal potential region.   Why do books also conclude, that the surface is at the same potential as well?
98037
If we have a point charge and outside of it we have a non-conducting Gaussian sphere, then Gauss's law says that the net flux should be zero. I agree that the total field lines coming in are equal to the number of field lines coming out, but isn't it true that the magnitude of the electric field at those segments of the sphere where the field is coming in is greater than those where it is coming out? So when we integrate, wouldn't these areas, having a greater electric field, contribute more to the net flux? I don't see why they would cancel each other out.
44940
This question is motivated by Section 3.2.3 in Griffiths. Therein, we are considering the force of attraction between a point charge and an infinite conducting plane.  One can calculate the field using the method of images, and from that the force of attraction exerted by the induced charge to be  $$\mathbf{F}=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}\widehat{\mathbf{z}}.$$  One one hand, the energy associated with a field is given by $$W=\frac{\epsilon_0}{2}\int\! E^2\ \mathrm{d}\tau.$$  Since the dot product is positive definite, this implies that $W$ must be positive.  On the other hand, the work is also given by \begin{align} W&amp;=\int_\infty^a\! (-\mathbf{F})\cdot \mathrm{d}\mathbf{l}\\ &amp;=\frac{1}{4\pi\epsilon_0}\int_\infty^a \frac{q^2}{4z^2}\ \mathrm{d}z\\ &amp;=-\frac{1}{4\pi\epsilon_0}\frac{q^2}{4d}&lt;0 \end{align} Intuitively, it makes sense that the work should be negative: to bring the particle in from infinity under constant acceleration, we must oppose the attractive force of the induced charge.  But I am having trouble reconciling this with the equation $W=(\epsilon_0/2)\int\! E^2\ \mathrm{d}\tau.$  I am sure I am missing something, and would greatly appreciate a nudge in the right direction.
27685
Let there be a hollow sphere (Throughout the question we will ignore the thickness of the sphere) which is positively uniformly charged,q of radius,  R. Suppose there to be a test positive charge, q' at point P which is r distance away from the hollow sphere(r>R, P is outside of hollow sphere).  Now I have an doubt on evaluating Electric Potential,V at point P.  Both source charge,q and test charge,q' are +ve, so source charge will repel the test charge which will make the charge, q' to go to infinity.I have to apply external force, F' to displace the charge, q' from infinity to point P against the equal but repulsive force, F.In short, F' is  displacing the q' from infinity to P.  So External Force, F' will do +ve work and Repulsive Force, F will do -ve work. This way the Electric Field,E produced by hollow sphere is Opposite to Displacement of test charge(i.e.,θ=180).  Now we know that E=k(q/(r^2))  [Please Notice that infinity is the lower limit of integration]  Also V=-∫r∞(E dr cosθ)=-∫r∞(E dr cos 180)=∫r∞(E dr)  At last I ended up to V=- k(q/r). I know that V=k(q/r).  I just want to know why I am having this shitty negative sign in my equation.
84745
The electric force (Coulomb's law) on a point charge $Q_2$ due to $Q_2$: \begin{gather*} \mathbf{F}_{12}=\frac{Q_1Q_2}{4\pi\epsilon_0}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3} \end{gather*}  The seperation vector between the charges is $\mathbf{r}_{12}=\mathbf{r}_2-\mathbf{r}_1=(x_2-x_1)\hat{\textbf{i}}+(y_2-y_1)\hat{\textbf{j}}=(z_2-z_1)\hat{\textbf{k}} $.  Does this mean that the force is a vector field or just a vector?   Is the force a function of $\mathbf{r}_{12}$, so $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?
88530
An  is defined as $p = q\times d$ (for two point charges $\pm q$ separated by a distance $d$).    What is the physical meaning of this quantity?  Why does the direction of the electric dipole moment go from the negative charge to the positive charge? Another way of saying this, why electric dipole moment is a vector quantity?