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16

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


9

In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...


4

In classical electrodynamics, assuming a point charge to be having a finite charge, the net electrostatic self energy carried by it is given by $$ Self Energy = 1/2 \int E^2 dV$$ Upon performing the intergral in three dimensions, since the electric field of a point charge diverges at the origin, therefore the rest mass by the virtue of the electrostatic ...


3

No, atoms have the same number of protons and electrons so they have no net charge. On the other hand ions (cations and anions) would be repelled or attracted depending on their net charge. Atoms are bound together in a molecules by different means like covalent bonding, ionic bonding (which can be easily explained in terms of electrostatic forces) or ...


2

Assuming incandescent bulbs rather than, say, LEDs: they won't take "almost all" the energy, they'll take all of it. It doesn't matter if you use one bulb or five, it'll use up all of the energy that the battery is supplying. That's why multiple bulbs will be dimmer than one single bulb. Electrons collectively don't have a very good sense of how much energy ...


2

The motion of the charges is totally unlike you setting off from home walking 10 km and feeling a little tired and slowing down and then walking another 10 km and feeling even more tired and slowing down and then walking a final 10 km and arriving home at a very slow pace, indeed just reaching your front door and stopping. Here is a simple model of what ...


2

There many configurations that statisfy your assumptions. But you had forgot about many other constraints like potential difference, charge density, end effects etc. When you consider PD you come to know, ignoring end effects, that only one configuration is possible. Since 1st plate has charge Q its surface charge density of 1st side is Q/2A 2nd side again ...


2

In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


1

Current constituted by positive ions doesn't imply that the current is constituted by protons. The positive charge might also be constituted by ions, or holes in semiconductors. If you are referring specifically to a circuit made up of solid material, then protons are tightly bound and hence don't flow and positive charge if any, is due to holes. Protons ...


1

You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what ...


1

Atoms are electrically neutral. Because of this they shouldn't attract or repel each other - but atoms do show a slight attraction, which is the reason most molecules form. This is called the residual electromagnetic interaction. In short, the positive parts of one atom attract the negative parts of the other, and vice versa. There is a good little diagram ...


1

Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct. First consider no cavity Potential at center of sphere due to uniformly charged complete sphere $ V = 3kq/2a$ Now, potential due to positive charged sphere $cavity$ at center. $$ V_{1}= \frac{4\rho π ...



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