Two point charges, 5.0 µC and -2.0 µC, are placed 5.8 cm apart on the x axis, su

Question

Two point charges, 5.0 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 5.0 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.)
x < 0 ____ cm
0 < x < 5.8 cm ____ cm
5.8 cm < x ____ cm

(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.)
x < 0 ____ cm
0 < x < 5.8 cm ____ cm
5.8 cm < x ____ cm

Explanation / Answer

a) for the net field Eres to be zero,
directions of individual fields should be opposite in direction and equal in magnitude
for 0 < x < 5.8 cm, Eres != 0 at any point because both the fields are in same direction.

for 5.8 cm < x, 5.0 µC > 2.0 µC and r1<r2,
the individual fields cannot have equal magnitude

for x<0, the condition for field to be zero q1/r1^2 = q2/r2^2
5/(5.8+r)^2 = 2/r^2
r = 9.98 cm
x = -9.98 cm

x < 0 ___-9.98_ cm
0 < x < 5.8 cm _0___ cm
5.8 cm < x __0__ cm

b)The two potential should be opposite in sign and equal in magnitude
for 0 < x < 5.8 cm, q1/r1 = q2/r2
5/(5.8-r) - 2/r = 0
r = 1.66 cm
x = 1.66 cm

for 5.8 cm < x, 5.0 µC > 2.0 µC and r1<r2,
5/r<2/(5.8+r)
Hence there won't be any point x>5.8 where potential is zero

for x<0, 5/(5.8+r) - 2/r = 0
5/(5.8+r)^2 = 2/r^2
r = 3.87 cm
x = -3.87 cm

x < 0 ___-3.87_ cm
0 < x < 5.8 cm _1.66___ cm
5.8 cm < x __0__ cm

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