Question

A 2.2 mm -diameter glass bead is positively charged. The potential difference between a point 2.0 mm from the bead and a point 4.4 mm from the bead is 500 V. What is the charge on the bead?

Answer 1

r = d/2 = 2.2/2 = 1.1 mm = 0.0011 m

Let Q is the charge on the bead

let d1 and d2 are the distances from center of the bead.

d1 = 1.1 + 2 = 3.1 mm = 0.0031m

d2 = 1.1 + 4.1 = 5.2 mm = 0.0052 m

V1 - V2 = k*Q/d1 - k*Q/d2

V1 -V2 = Q*k*(1/d1 - 1/d2)

Q = (V1-V2)/(k*(1/d1 - 1/d2))

= 500/(9*10^9*(1/0.0031 - 1/0.0052))

= 4.26*10^-10 C or 0.426 nC