# Electricity and Magnetism (2-205)

**Physics 205, Electricity and magnetism. 2012, Problem set #2.**

1: The electric potential of a very large isolated flat metal plate is 2000 V. The plate carries a uniform distribution of charge with surface density $s=1 mC/m^2$. Determine potential V at a distance $x=1 cm$ from the plate. Assume that point x is far from the edges and that x is much smaller than the size of the plate.

2: Four charges, $q1=+q$, $q2=-q$, $q3=+q$ and $q4=-q$ are at the corners of a square with the side $a=7.5 cm$. (In such way that if one travels along the perimeter of the square, the charge signs are alternating.) If $q=3.0 mC$, what is the total energy required to assemble this system of charges? Comment on the sign of your answer.

3. The metallic sphere on top of a large van de Graaf generator has a radius of 2.0 m. Suppose that the sphere carries a charge of $3*10^-5 C$. How much energy is stored in this device?

4. The HCl (hydrochloric acid) molecule has a dipole moment of about $3.4*10^-30 C*m$. The two atoms are separated by about $1*10^{-10} m$. What is the net charge on each atom? What is maximal torque that this dipole will experience in a $5.104 N/C$ electric field? How much energy one would need to rotate one molecule $45 degree$ from its equilibrium orientation?

5. An isolated charged parallel-plate capacitor with the plate separation $d=2 mm$ and area of $A=10 cm^2$ is initially empty. One slides a polystyrene dielectric slab with thickness equal to half the plate separation and with only half the area of the plates into the region between the plates. What is the final capacitance?

Answers

1. Since the plate is very large the filed at any point is the same. As the Gauss law states

$E = \sigma/(2*\epsilon_0)$, where epsilon is the electric permittivity of the air.

Now the relation between the potential and the electric field at point x is

$\delta(V) = – E*x = – \sigma*x/2/\epsilon_0$

where delta is the change in the potential.

$V = V_0 – \sigma*x/2/\epsilon_0 = 2000 – 10^{-6}*0.01/2/8.85*10^{-12} = 1435 V$

2. The potential energy is a sum of individual pairs potential energies.

$V = 1/(4*\pi*\epsilon_0) (q_1*q_2/R_{12} + q_1q_3/R_{13} +q_1q_4/R_{14} + q_2q_3/R_{23} +q_2q_4/R_{24} +q_3q_4/R_{34}) =$

$=(q^2/4\pi*\epsilon_0)*(-1/a +1/a*\sqrt{2} -1/a -1/a +1/a*\sqrt{2} -1/a) =$

$=9*10^{-12}*9*10^9 *(-3/0.075 +2/0.075*1.41) = – 81*10^{-3}*21.08 =1.708 J$

3. It is like all the charge is located at the center of the sphere with radius R. The potential is

$V = q/(4*\pi*\epsilon_0*R) = 3*10^{-5} *9*10^9/2 =135000 V$

The energy stored is

$W =e*V =1.6*10^{-19}*135000 =2.16*10^{-14} J$

4. The dipole is composed of charges +q and -q at the distance x. Hence the dipole p is

$p =q*x$,

$q =p/x =3.4*10^{-30}/10^{-10} =3.4*10^{-20} C$

the torque in and electric filed E is simply

**T = p** x **E** where x is the vector product so that the maximum torque is

$T = 3.4*10^{-20}*5*10^4 =1.7*10^{-15} N*m$

The potential energy of a dipole p in an electric field E is

$U = -p*E$ where * is the scalar product between the two vectors

Choosing U =0 at the angle zero one has

$U =3.4*10^{-20}*5*10^4*cos(45) = 1.2*10^{-15} J$

5. After sliding the polystyrene inside there are three capacitors connected is series with the distance between plates half of the initial distance. Initial capacitance

$C_0= \epsilon_0*A/d =8.85*10^{-12}*10*10^{-4}/0.002 = 4.425 pF$

The capacitor having dielectric only air and distance between plates initial d has capacitance and half plates area has capacitance

$C_1 = C_0/2 = 2.21 pF$

The capacitor having dielectric air and having distance between plates 1 mm and half area

$C_2 = C_0*2/2 = 4.425 pF$

The capacitor having dielectric polystyrene and having distance between plates 1 mm and half area has capacitance

$C_3 =\epsilon(polystyrene)*C_2 = \epsilon(polystyrene)*4.425 =2.5*4.425 =11 pF$

The total capacitance is

$C = C_1 parallel (C_2 series C_3) = 2.21 +(4.425*11)/(4.425+11) =2.21+3.15 = 5.366 pF$