Answers to homework problems

Chapter 19
  2) 
     a). The force acting on the proton is equal to 
         F = E*q = 2.0x10^4 N/C * 1.6*10^(-19) C = 3.2*10^(-15) N 
         The direction of the force is the same as that of the field and is 
         in the direction of the movement. Since we can measure only the 
         difference between potential energies, let us choose EPE at point A
         as EPEa = 0. The change in the proton's electric potential energy is 
         EPEa - EPEb = 0 - F*d = -3.2x10^(-15) N * 0.15m = -4.8*10^(-16) J. 
     b). In that case the direction of the force is opposite to the direction 
         of the proton's movement. So EPEa - EPEb = 0 - (-F)*d = 4.8*10^(-16)J

  3) Work done by the electric force on the electrons which move from one
     potential to another is equal to the potential difference times the charge.
              W = V * q
     and the total charge is equal to the charge of a single electron
     multiplied by the number of electrons
               q = q_0 * N
     Substitution gives
              W = V * q_0 * N
     or
                      W      
              N = ------- 
                   V * q_0  
     V*q = 20000 V * 1 e = 20000 eV, W = 1.5x10^(-7)J = 9.3x10^11 eV, so

                      1.5 * 10^-7 J
              N = ----------------------- = 4.7 x 10^7 = ~47 million electrons!
                    20000V * 1.6 10^-19 C

 12) Electric potential of the point like-charge can be written in terms 
     of distance r, and the charge q as:
              V = k*q/r
     V= 164 V at r = 0.20 m so the charge is q=V*r/k
     q = 164 V * 0.2 m/8.99 10^9 Nm^2/C^2 = 3.65 10^-9 C
     The potential at distance 0.8 m is
              V = k*q/r =8.99 10^9 Nm^2/C^2 * 3.65 10^-9 C / 0.8 m = 41 V

 16) Potential energy of the system of the two point charges is
              W = k * q1 * q2 / r
     where q1 -- is a charge of one particle, q2 -- is the charge
     of another particle, r -- the distance between them.
     When the charges are moved there is a  new potential 
       W_1 = k * q1 * q2 / r_1,
     where r_1 is the new distance between charges. 
     Since W_1 = 2W 
     K * q1 * q2 / r = 2 * k * q1 * q2 /r_1  ---> r_1 = 2 * r

 19) We need to consider two charges picture placed on the diagonal
     of the square and a thrid one at its center. If the potential 
     at one of the empty corners is zero, it will also be zero at the other
     empty corner due to the symmetry. Potential of the empty corner is 
     
     V = k * q_1/ r_1 + k * q_2/ r_2 + k * q_3/ r_3 
       = k * (q_1/r_1 + q_2/r_2 + q_3/r_3)
   
     The charges are q_1= +9q, q_2 = -8q and q_3 = x (unknown).
     The distances are r_1 = r_2 = a (side of the square) and
     r_3 = sqrt(2) * a / 2 (half of the diagonal of the square)
     The potential is
      V = k * ( 9q/a -8q/a + x/(sqrt(2)*a/2) = 0
      q/a + x/sqrt(2)*a/2 = 0
      x = - q * sqrt(2)/2 = -q /sqrt(2) (negative charge)
   

 31) Potential V can be written in terms of
     electric field E and distance between points r as
     V = E * d 
     V = 2.8 * 10^6 V/m * 7.5 * 10^(-4)m = 2100 V 

 40) Capacitance of a parallel plate capacitor is  
       C = e_0 * A / d
     A -- is the area of each plate , d -- is a distance
     between them and e_0= 8.85 10^-12 C^2/(nm^2)
     A = d * C / e0 
       = 1 m * 1 F / (8.85*10^(-12) C^2/Nm^2) = 1.13*10^11m^2 =
       ~43,600 square miles