Question

A -12.7nC point charge and a +21.3nC point charge are 13.8cm apart on the x-axis. What is the electric potential at the point on the x-axis where the electric field is zero?

Incompatible units. No conversion found between "m/s" and the required units.

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What is the magnitude of the electric field at the two points on the x-axis where the electric potential is zero? (Input your answers in order of increasing distance from the negative point charge.)

Answer 1

Here, for the electric field to be zero,

the position is left of -12.7 nC at x

for zero field

k * 12.7/x^2 = k * 21.3/(x + 13.8)^2

12.7/x^2 = 21.3/(x + 13.8)^2

solving for x

x = 46.8 cm

potential at this point = 9 *10^9 * 10^-9 * (-12.7/0.468 + 21.3/(0.138 + 0.468))

potential at this point = 72.1 V

the potential at the point where field is zero is 72.1 V

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Now, for the potential to be zero

in between charges

k * 12.7/x = k * 21.3/(13.8 - x)

12.7/x = 21.3/(13.8 - x)

x = 5.15 cm = 0.0515 m

in left of charges

k * 12.7/x = k * 21.3/(13.8 + x)

12.7/x = 21.3/(13.8 + x)

solving for x

x = 20.4 cm = 0.204 m

Now, for the electric field at first point

E = 9 *10^9 * 10^-9 * (12.7/0.0515^2 + 21.3/(0.138 - 0.0515)^2)

E = 68716 V/m

for the electric field at the second point

E = 9 *10^9 * 10^-9 * (12.7/0.204^2 - 21.3/(0.204 + 0.138)^2)

E = 1107.6 V/m

the electric field at the points are 68716 V/m and 1107.6 V/m