Two spheres, one with charge Q_1 = -3 Times 106-8 C and the other with charge Q_

Question

Two spheres, one with charge Q_1 = -3 Times 106-8 C and the other with charge Q_2 = 5 Times 10^-8 C, are located 3m apart, as shown in the figure below. Point A is 1m from Q_1 and 2m from Q_2, and point B is 4m from Q_1 and 5m from Q_2. The angle Q_1-B-Q_2 is a right angle, and the angle Q_1-Q_2-B is 36.9 degree. What is the electric field at point A. and what is the electric field at point B? (Give your answers as either magnitudes and directions or in terms of x and y components.) What is the potential difference between points A and B?

Explanation / Answer

A) Electric field = kq/d^2

At point A

E(at point A) = k ( Q1 ) / d1^2 + k ( Q2 ) / d2^2

Notice that this happens because the field from Q1 is towards Q1 (the negative direction), and the field from Q2 is also away from Q2 in negative direction.

k = 9e9

d1 = 1m

d2 = 2m

EA = 382.5 N/C

Now at point B

Electric field due to Q1 is towards Q1 ie in positive y direction

EQ1y = kQ1/d1^2 = 9e9*3e-8 / 4^2 = 16.875 N/C

EQ1x = 0

Electric field due to Q2 is away from Q2 ie in negative x-y direction

EQ2x = - kQ2/d2^2 * sin(36.9) = - 10.8 N/C

EQ2y = - kQ2/d2^2 * cos(36.9) = - 14.4 N/C

Ex = -10.8 N/C

Ey = 16.875 - 14.4 = 2.475 N/C

Hence EB = sqrt((-10.8)^2 + 2.475^2) = 11 N/C

B) Electric Potential is a scalar quantity

VA =  k ( Q1 ) / d1 + k ( Q2 ) / d2 = -45 V [d1=1, d2=2]

VB = k ( Q1 ) / d1 + k ( Q2 ) / d2 = 22.5 V [d1=4, d2=5]

VB - VA = 22.5 - (-45) = 67.5 V

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