Two point charges, 6.0 µC and -2.0 µC, are placed 5.8 cm apart on the x axis, su

Question

Two point charges, 6.0 µC and -2.0 µC, are placed 5.8 cm apart on the x axis, such that the -2.0 µC charge is at x = 0 and the 6.0 µC charge is at x = 5.8 cm.

(a) At what point(s) along the x axis is the electric field zero? (If there is no point where E = 0 in a region, enter "0" in that box.)

(b) At what point(s) along the x axis is the potential zero? Let V = 0 at r = . (If there is no point where V = 0 in a region, enter "0" in that box.)

Explanation / Answer

a)

q1 = - 2 x 10-6 C

q2 = 6 x 10-6 C

r = distance between the charges = 5.8 cm = 0.058 m

d = distance of charge q1 from zero electric field point

r + d = distance of charge q2 from zero electric field point

at zero electric field point

E1 = E2

k q1/d2 = k q2/(r + d)2

(2 x 10-6 )/d2 = (6 x 10-6 )/(0.058 + d)2

d = - 0.021 or 0.079 m

d = - 2.1 or 7.9 cm

b)

at zero electric potential point between the charge particles

V1 + V2 = 0

k q1/d + k q2/(r - d) = 0

(- 2 x 10-6 )/d + (6 x 10-6 )/(0.058 - d) = 0

d = 0.0145 m = 1.45 cm

at zero electric potential point not between the charge particles and closed to charge q1

V1 + V2 = 0

k q1/d + k q2/(r + d) = 0

(- 2 x 10-6 )/d + (6 x 10-6 )/(0.058 + d) = 0

d = 0.029 or 2.9 cm

x < 0 ,   at x = - 2.9 cm

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