Numerical-4

Q .  Two point charges q1 and q2 , of magnitude +10^–8 C and 

–10^–8 C, respectively, are placed 0.1 m apart. Calculate the electric

fields at points A, B and C shown in Fig. 1.14.

Solution : 

The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude

E(1A) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.05 m)^2  = 3.6 × 10^4 N C–1 

The electric field vector E(2A) at A due to the negative charge q2 points towards the right and has the same magnitude. 

Hence the magnitude of the total electric field E(A)  at A is

E(A)  = E(1A) + E(2A) = 7.2 × 10^4 N C–1

E(A)  is directed toward the right.

The electric field vector E(1B) at B due to the positive charge q1 points towards the left and has a magnitude

E(1B) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.05 m)^2  = 3.6 × 10^4 N C–1 

The electric field vector E(2B) at B due to the negative charge q2 points towards the right and has a magnitude

E(2B) =  (9 x 10 Nm 2 C-2 ) x (10^8 C) / (0.15 m)^2  = 4 × 10^3 N C–1 

The magnitude of the total electric field at B is

E(B) = E(1B) – E(2B) = 3.2 × 10^4 N C–1

E(B)  is directed towards the left.

The magnitude of each electric field vector at point C, due to charge

q1 and q2 is

E(1C) - E(2C)=  (9 x10 Nm 2 C-2 ) x (10^8 C) / (0.10 m)^2  = 9 × 10^3 N C–1 

The directions in which these two vectors point are indicated in

Fig. 1.14. The resultant of these two vectors is

E(C) = E(1) cos π/3 + E(2) cos π/3 = 9 × 10^3 N C–1

E(C)  points towards the right.