Question

Two point charges lie on the x-axis. A charge of -5.5 microcoulombs is at the origin, and a charge of +4.5 microcoulombs is at x = 10.0 cm.

a. what is the net electric field at x = -4.0 cm?

b. what is the net electric field at x = +4.0 cm?

Answer 1

We know that

E = kQ/R^2

And direction of electric field is away from positive charge and towards negative charge.

Part A.

At x = -4 cm

due to Q1, E will be in +ve x-axis and due to Q2, E will be in -ve x-axis

Enet = E1 - E2

Enet = kQ1/R1^2 - kQ2/R2^2

R1 = 4 cm & R2 = 14 cm

Enet = 9*10^9*5.5*10^-6/0.04^2 - 9*10^9*4.5*10^-6/0.14^2

Enet = 28.87*10^6 N/C toward +ve axis

Part B.

At x = +4 cm

due to Q1, E will be in -ve x-axis and due to Q2, E will be in -ve x-axis

Enet = - E1 - E2

Enet = -kQ1/R1^2 - kQ2/R2^2

R1 = 4 cm & R2 = 6 cm

Enet = -9*10^9*5.5*10^-6/0.04^2 - 9*10^9*4.5*10^-6/0.06^2

Enet = -42.19*10^6 N/C

Enet = 42.19*10^6 N/C toward -ve axis

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