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Chapter 4: Q81P (page 718)

A negative point charge q1= -4.00 nC is on the x-axis at x = 0.60 m. A second point charge q2is on the x-axis at x = -1.20 m. What must the sign and magnitude q2be for the net electric field at the origin to be

(a) 50.0 N/C in the +x direction and

(b) 50.0 N/C in the 鈥搙-direction?

Short Answer

Expert verified

a) q2is -8.0nC when the net electric field at the origin is 50.0 N/C in the +x direction.

b) q2is -24.0nC when the net electric field at the origin is 50.0 N/C in the +x direction

Step by step solution

01

Calculation of charge in +x direction.

Given r1=0.6m,r2=1.2mand

q=-4nc

a) The net electric field at the origin of the first charge is positive x is equal to

E1=kq1r21=9脳10^9脳4脳10-9(0.6)2=100N/C (1)

so the electric field due to the second charge must be -x and its magnitude must be 50N/C so the second charge is negative and is equal to

E1=kq1r21=q=50脳1.229脳109=8.0脳10-9c8.0nC

So the charge is q2=-8nC
02

Calculation of charge in -x direction.

The net electric field at the origin of the first charge is positive x and is equal to

E1=kq1r21=9脳10^9脳4脳10-90.62=100N/C (1)

so the electric field due to the second charge must be -x and its magnitude must be 150N/C so the second charge is negative and its magnitude is given by

q=50脳1.229脳109=8.0脳10-9c8.0nC

So the charge is q2= - 24nC

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