1. The change in potential (denoted by V) is - 100 Volts - 0 Volts = -100 Volts and the charge is negative. Consider two different arguments.

• Since the potential got smaller (the particle moved in the direction of the potential drop) ===> the particle moved in the direction of the field. Since the particle has a negative charge, if left alone, it would have moved in the opposite direction.

• The potential energy is PE = qV and q less than 0 ==> the change in PE is greater than 0, that is, the PE increased; the particle must have been "pushed" by an external force to raise its potential.

2. Inside of the hollow, spherical conductor the electric field is 0. This means that the potential V must be constant for r less than R. Suppose that it has a value of Vin (What is Vin? We will answer that in a second). Outside of the conducting sphere, the potential falls off as kQ/r. Since the potential must be continous at the surface of the shell, r = R, the potential for r less than R is Vin = Vout(r=R) = kQ/R.

3. The charged spherical conductor will induce a charge separation in the nearby neutral conductor pulling negative charges closer to it leaving the far side positively charged

Now consider how the potential changes. In our discussion we start near the charged conductor and move along the line connecting the two spheres. The negative charges on the neutral sphere, being closer to the charged sphere than are the positive charges of the neutral sphere, lower the potential in the interior of the charged sphere and close to the charged sphere. The potential levels out as it passes through the neutral conductor. Beyond the neutral conductor, the potential larger than for the case without the neutral conductor due to the contribution of the positive charges on the neutral conductor.

4. Grounding the neutral conductor will allow negative charges to be "pulled" from the Earth onto the conductor neutralizing it. In this case, the potential is strongly reduced everywhere.

13. All potential measurements only detect changes in the potential. So, if we perform measurements on the surface of the Earth, whether the Earth has 0 potential or a huge potential is irrelevant. The Earth makes the same contribution to all potentials, that is, we can consider the total potential as the potential to the charges (we are interested in and the potential of the Earth and so,

Changes in potential = (v2+v[Earth]) - (v1+v[Earth]) = v2 - v1

Problems

1. The work is Work = change in PE = q x change in V = -25 x 10**(-9) C x 100 V = -2.5 x 10**(-6) Joules. The negative sign means that the potential energy got smaller. This means, from conservation of energy, that the particle's kinetic energy (KE) got larger. The field did work on the charge accelerating it to higher KE.

3. The direction of the field was not specified. Let the left hand plate be placed t x = 0 cm and the right hand plate be at x = 10 cm. Now suppose that the field points in the direction of increasing x.

E = -(change in V) / d ===> V(x=10cm) - V(x=0 cm) = -1 Volts/m x 0.1 m = -0.1 Volts

5. The inner sphere induces a charge of -Q on the inner surface and a charge +Q on the outer surface of the outer conducting shell. Since the outer sphere is grounded, however, the charge on the outer surface of the conducting shell is neutralized and so, we see only the +Q of the inner sphere and the -Q of the inner surface of the outer shell. The total charge is 0 and the potential is then V = 0 for r greater than R.

7. If the spheres are very small, then the potential will be the sum of the potentials of the individual spheres. That is,

V = k (30 x 10**(-6) C)/0.5 m - k (50 x 10**(-6) C)/0.5 cm
===> V = - 3.6 x 10**(5) Volts

10. The energy gained by the alpha particle is 100 kilo-electron Volts = 100,000 q(e) Volts which must be equal to the work performed on it by the field. The gain in KE = Work or 10**5 q(e) Volts = 2 x q(e) x V ===> V difference = 50,000 Volts

17. To suspend an electron in "midair" means that we must balance the downward pull of gravity. Therefore

-m(e) g - q(e) E = -m(e) g + q(e) (change in V) / d = 0
===> change in V = [m(e)gd]/q(e) = 5.6x10**(-12) Volts

18. By symmetry,

• the electric field will be zero at the center of the square.
• The potential will be the sum of the potentials of the four charges. By symmetry, the potentials for all 4 charges are the same. So, we have
  V = 4 x k [60x10**(-6) C] / sqrt(0.01+0.01) = 15 million Volts

20. The potential is the sum of the four potentials. The total potential is then turns out to be

V = [k/sqrt(0.08)](sum of the charges) = 0 Volts.
For fun, what is the electric field at the center of the square? Will it also be 0?