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You have ignored the mobile charges in the conductor. In your plot the field lines are not perpendicular to the surface, particularly near the charges. That will cause the conduction electrons to move. The positive charges will attract electrons until the field inside the conductor is zero. This means that the whole conductor, including the inner ...

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...why isn't the work done against the net force due to the system considered instead of simply adding up the work done against separate forces caused by individual charges? They're both equivalent, due to the principle of superposition. Basically, the net force is what you get when you add up the separate forces from the individual charges acting on ...

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In the case of a grounded conducting shell, it is well known that the method of images can be used to calculate how the total charge $-q$ on the inner surface is distributed. The solution is given in the wikipedia link above. Now, you ask what happens if the the potential of the shell is fixed, but not necessarily zero. The electric field inside the ...

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We can easily do this calculation. The capacitance of a sphere is: $$C = 4\pi\varepsilon_0r$$ and the charge is given by: $$Q = CV = 4\pi\varepsilon_0r V$$ The number of extra electrons is: $$n_e = \frac{Q}{e} = \frac {4\pi\varepsilon_0r V}{e}$$ And finally the mass of the extra electrons is: $$M = m_e n_e = \frac{Q}{e} = \frac ... 3 You have a sign error. The potential is defined by see: http://en.wikipedia.org/wiki/Electric_potential 3 Very simply, the field of the positive and negative elements of the dipole "almost" cancel out - but not quite. It is because they are some small distance away that there is a residual (third order) term. You can see this by taking two charges +q and -q at a distance 2d, and look at the field a distance r from the center of the two (on the same ... 3 The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ... 3 Consider the Earth-Moon system. They are subject to an attractive force (gravitation) and to no repulsive forces (neglecting solar tides, anyway), yet they stay at a nearly constant distance from one another because of their dynamics. A a static analysis of this system would prompt us to postulate some repulsive force holding the bodies apart (and you can ... 3 When you ask questions about things "in the limit", the answer is almost always "It depends". In this case, the answer is "it depends". The equation Q=CV assumes linear behavior of the capacitor - in reality the dielectric of most capacitors has hysteresis as well as a nonlinear component, so as you increase the voltage, the capacitance will change. This ... 3 Your expression for the electric field in Cartesian coordinates is incorrect. The field is not radial, it points away from the line of charge. If we put the line of charge along the z-axis, then the E-field is$$\vec{E} = \frac{x\vec{i} + y\vec{j}}{x^2 + y^2}with some multiplicative constant involving the charge per unit length and \epsilon_0. The ... 2 A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be \mu_0 \vec{J}, where \vec{J} is the current density. If the ... 2 You can't take the root mean square of the spherical coordinate parameters because they aren't all the same units (one is a length measurement while the other two are angle values). Well you can, but the output is meaningless. To convert spherical parameters to Cartesian coordinates, you use simple trig: \begin{align}E_z & = r \cos(\theta) \\ E_x & ... 2 If the shell and its charge distribution are spherically symmetric and static, and if electric field lines begin and end on charges, then we know that any electric field that might be present inside the shell must be directed radially (in or out). From there, a simple application of Gauss's law, using a spherical surface centered on the center of the shell ... 2 The boundary conditions you mention,\vec{D} = \rho_s\hat{a}_z$$are for charges distributed on the surface of a volume conductor, following from the requirement that the electric field is zero inside the conductor. The expression$$\vec{E} = \frac{\rho_s}{2\epsilon}\hat{a}_z$$is instead valid for a planar sheet of a conductor alone (or for that matter, ... 2 Yes, you can use the method of images because uniqueness of the solution is guaranteed when you Know the total charge of an equipotential surface without knowing the value of the potential itself I'll summarize a procedure to obtain the correct answer: Application of Gauss law tells us that there must be total charge -q on the inner surface then because of ... 2 Answering your three questions: He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 µm is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size ... 2 In the context of ion beams, space charge is the tendency of the beam to expand transversely (perpendicular to the direction of the beam's travel) due to the mutual repulsion of the ions in the beam. All the ions have the same sign charge, so they repel. The name "space charge" comes from plasma physics where is is often computationally easier to treat the ... 2 Your first two equations are effectively the same once you consider that F = q E. You should use the second one if you're concerned about the electric field in general, but the first one if you want to know what actually happens to a particle (i.e., what force a particle feels). When in doubt, you can usually do alright by working out first the field, and ... 2 You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't. In the case of a switched AC outlet, the switch will be ... 2 Electrons flow through the wire from the battery's negative terminal to the battery's positive terminal. If said wire is actually a capacitor, the electrons still flow the same way - from the battery's negative terminal. But since it's a capacitor, the electrons are pushed into the capacitor's negative plate instead of making it all the way to the battery's ... 2 Basically, the system is simply that which is studied in a problem in physics. It refers to that which we want to know more about, in this case the moving electric charge in the presence of the electric field. Be cautious with the terms 'electric potential' and 'potential energy', since they're two different things. Electric potential is defined as ... 2 Observe the potential lines for a moment. You will find that for equal change in distance, there is equal change in potential. Means, if I move 0.5 m to the left, the potential increase is 10 V. In other words, we have equidistant equipotential lines which is a graphical way of denoting uniform field. Whenever you see straight equipotential lines, it means ... 2 Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small ... 2 A negatively charged particle has an electric field,$$\mathbf{E} =-\vert\mathbf{E}\vert \, \hat{\mathbf{r}} =\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}Gauss's law gives, \begin{align*} \int \mathbf{E}\cdot d\mathbf{A} &= \int \Big(\frac{-q}{4\pi\epsilon_0 r^2}\hat{\mathbf{r}}\Big) \cdot (r^2 \sin \theta d\theta d\phi \, \hat{\mathbf{r}}) = ... 2 It depends on what you mean when you say V=0. In the context of the equation:\vec{E}=-\nabla V$$which holds specifically in electrostatics V is a scalar field, meaning that it is actually a function which assigns every point in space a scalar value. \vec{E} is a vector field, which assigns a vector to every point in space. Thus, both the electric ... 2 In electrostatics, when two conductors are connected to each other, their charges redistribute such that their potential difference is minimized. (Ideally the difference should be zero.) This is the most stable state for a system of conductors. Theoretically, since the Earth is usually taken to be at zero potential, a grounded conducting plate would be at ... 1 "So if I construct a gaussian surface such that it extends out from both the sides ,then I should get the same answer as that for a charged sheet! Why do we not do this in this case ?" if you do that you have to realize that the other side have same amount of charge as the first, so the filed is doubled As for your last question if you consider a ... 1 Because they should have the same electric potential and electric potentials of them depends on charge not charge density. 1 The potential at the surface of a charged sphere or radius r is:$$ V = k\frac{Q}{r} \tag{1} $$Since the area of the sphere is 4\pi r^2, the charge density is:$$ \rho = \frac{Q}{4\pi r^2} $$and rearranging gives:$$ Q = \rho 4\pi r^2 $$and substituting for Q in equation (1) we get:$$ V = 4\pi k\rho r  Can you take it from here?

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I will presume that, as is standard and conventional, the electric potential at a place an infinite distance from the one in consideration is zero. Let the charge at point a be charge $a$, the charge at point b be charge $b$ and the charge at point c be charge $c$. With this, the formula for the electric potential is: \$V = ...

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