1. For the charge distribution shown in figure 1. a) Find the field at the origi

Question

1. For the charge distribution shown in figure 1. a) Find the field at the origin Find the potential at the origin b) c) figure 1 What work will you have to do on a 3C charge to move it from infinity to the origin? For the charge distribution shown in figure 2. a) Find the field at the origin b) Find the potential at the origin c) What work will you have to do on a -2C charge to move it from infinity to the origin? 2. -Q figure 2 figure 3 For the charge distribution shown in figure 3. a) 3. b) c) Find the field at the origin Find the potential at the origin What work will you have to do on a 3C charge to move it from infinity to the origin?

Explanation / Answer

1.a electric field E =Kq/r^2

   electric field at a point due to four charges q1,q2,q3,q4 at distance r1,r2,r3,r4 is given by

E =E1+E2+E3+E4

q1=Q r1 =2a q2 =-Q r2=a q3 =Q r3=a q4 =-Q r4 =2a

   E =K( q1/r1^2+q2/r2^2+q3/r3^2+q4/r4^2)

E = K(Q/4a^2+(-Q)/a^2+Q/a^2+(-Q)/4a^2)

   field at origin E =0

b. Potential at origin is also zero as there is no field . As Electric potential V=E.d

E=0 ,    V=E.d =0.(d) =0

c. As there is no electric potential at origin due to the charges work done W =q* V, given q=3c

V= 0 therfore W =(3c)(0) =0

work done on the chrge 3c to move from infinity to origin is zero.

2.a) given q1 =Q , r1 =2a , q2 =-Q , r2= a , q3 = -Q , r3= a , q4 = Q , r4 = 2a

   E =E1+E+E3+E4

   E = K( q1/r1^2+q2/r2^2+q3/r3^2+q4/r4^2)

   E=K(Q/4a^2+(-Q)/a^2+(-Q)/a^2+Q/4a^2)

   E= K(Q/2a^2 - 2Q/a^2)

   E=K(-3Q/a^2) = -3Q/a^2 N/c

b. Electric potential V = Kq/r

   V=V1+V2+V3+V4

   V = K( q1/r1+q2/r2+q3/r3+q4/r4)

V= K( Q/2a+(-Q)/a+(-Q)/a+Q/2a)

   V = K(-Q/a)

   V = -KQ/a volts

c) work done in moving charge -2c from infinity to origin W =q * V

   q= -2C V =-KQ/a

W =q*V = (-2C) (-KQ/a) =2KQC/a joules

3a)    q1=Q r1 =a q2 =-Q r2= a

   As the Q is along y axis field due Q be Ey =Kq/a^2

As the -Q is along X axis field due to -Q be Ex = -KQ/a^2

Resultant of two electric field vectors =SQRT( Ex^2+Ey^2)

   Er = SQRT((KQ/a^2)^2 +(-KQ/a^2)^2)

Er =1.414KQ/a^2 N/C

b) electric potential V= SQRT(Vx^2+Vy^2) where Vx, Vy are horizontal and vertical components of electric potential

Vx =-KQ/a

   Vy= KQ/a

   V = SQRT((KQ/a)^2+(-KQ/a)^2)

   V = 1.414KQ/a volts

3. work done in moving 3c charge fom infinity to origin W =q*V

given q=3C V=1.414 KQ/a

   W =(3c)(1.414KQ/a)= 4.242 KQC/a joules

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