Question

A charge of -2.705 μC is located at (2.830 m , 4.456 m ), and a charge of 1.460 μC is located at (-2.757 m , 0).

Find the electric potential at the origin. the answer to this was 154 V.

Here is the part I dont understand. Need (x,y) coordinates

there is one point on the line connecting these two charges where the potential is zero. Find this point.(x,y)

Express your answers using three decimal places separated by a comma.

Please show work and answer in (x,y)

Answer 1

given,

charge of magnitude -2.705 * 10^-6 C is located at (2.830 , 4.456) m

charge of magnitude 1.460 * 10^-6 is located at (-2.757, 0) m

electric potential = kQ / r

distance of first charge from the origin = sqrt((2.830 - 0)^2 + (4.456 - 0)^2)

distance of first charge from the origin = 5.28 m

distance of second charge from the origin = sqrt((-2.757 - 0)^2 + (0 - 0)^2)

distance of second charge from the origin = 2.757 m

electric potential at origin due to both the charges are = 9 * 10^9 * (-2.705 * 10^-6) / 5.28 + 9 * 10^9 * (1.460 * 10^-6) / 2.757

electric potential at origin due to both the charges are = 155.255 V

distance between these two charges = sqrt((-2.757 - 2.83)^2 + (0 - 4.456)^2)

distance between these two charges = 7.146 m

if the total potential is 0 at distance x fom second charge

so,

9 * 10^9 * (-2.705 * 10^-6) / (7.146 - x) + 9 * 10^9 * (1.460 * 10^-6) / x = 0

x = 2.51 m

since angle = tan^-1((y2 - y1) / (x2 - x1))

angle = tan^-1((0 - 4.456) / (-2.757 - 2.830))

angle = 38.57 degree

y co ordinate = 2.51 * sin(38.57)

y co ordinate = 1.565 m

x co ordinate = 2.51 * cos(38.57)

x co ordinate = -1.962

point on the line connecting these two charges where the potential is zero is (-1.962, 1.565)