Question

An isolated point charge is placed somewhere on the x-axis. At the point x = 5.00 cm and y = 0, the electric field points in the positive x direction and has magnitude E = 10.0 N/C. At the point x = 10.0 cm and y = 0, the electric field points in the positive x direction and has magnitude E = 15.0 N/C.

a)

Refer to the previous problem. Where is the charge located?

This time, give the x coordinate in CENTIMETERS.

b)Refer to the previous problem. What is the magnitude of the charge?

Answer 1

Origin is (0,0)

The charge is at a point (X,0) , where X has to be found

E1 = Field at point ( 0.05m,0) = 10 N/C

E2 = Field at point ( 0.10 ,0) m = 15 N/c

10 N/C = kq/ (X-0.05)^2

15N/C = kq/(X-0.10)^2

10 x ( X-0.05 )^2 = 15 x ( X- 0.10)^2

Simplifying

X^2-0.40 X +0.0250 =0

Solving for X

we get X1= 0.322 m and X2= 0.0775 m

Obviously X2 is wrong solution , since in that case the field would be pointing to left at one location and to right at the other

Hence Charge would be placed at ( 32.2,0) cm with respect to origin

10 N/c = kQ / ( 0.322 - 0.05)^2

q = 10N/C x ( 0.322-0.05)m^2 / 9.0 x 10^9 Nt-m^2/C^2 = 8.22 x 10^(-11) Columbs - this charge has to be negative since field is towards +x-axis direction at both locations.