Question

Two charges are placed on the x axis. One of the charges (q₁ = +6.34C) is at x₁ = +3.00 cm and the other (q₂ = -20.7C) is at x₂ = +9.00 cm. Find the net electric field (magnitude and direction given as a plus or minus sign) at (a) x = 0 cm and (b) x = +6.00 cm.

Answer 1

here,

q1 = 6.34 C is at x1 = 3 cm = 0.03 m

q2 = - 20.7 C is at x2 = 9 cm = 0.09 m

a)

the net eletric field at x= 0 cm , Ea = - K * q1 /x1^2 + K * q2 /x2^2

Ea = 9 * 10^9 * (- 6.34 /0.03^2 + 20.7 /0.09^2) N/C

Ea = - 4.04 * 10^13 N/C

the net electric field at origin has magnitude 4.04 * 10^13 N/C and in negative x-direction

b)

the net eletric field at x= 6 cm = 0.06 m , Eb = K * q1 /(x - x1)^2 + K * q2 /(x2 - x)^2

Eb = 9 * 10^9 * (6.34 /0.03^2 + 20.7 /0.03^2) N/C

Eb = 2.7 * 10^14 N/C

the net electric field at origin has magnitude 2.7 * 10^14 N/C and in positive x-direction