A +4.0- μC charge is placed on the x axis at x = +3.0 m, and a −2.0- μC charge is located on the y axis at y = −1.0 m. Point A is on the y axis at y = +4.0 m. Determine the electric potential at point A (relative to zero at the origin).

9.6kV

Explanation:

Electric potential is given as

V = k*Q/R

Then the Net electric potential at point A due to given two charges will be:

V_net = V1 + V2

V_net = k*q1/r1 + k*q2/r2

q1 = +4.0*10^-6 C & q2 = -2.0*10^-6 C

r1 = distance between q1 and point A = sqrt (3.0^2 + 4.0^2) = 5.0 m

r2 = distance between q2 and point A = 4.0 - (-1.0) = 5.0 m

Thus,

V_net = k*(q1/r1 + q2/r2)

V_net = 9*10^9*(4.0*10^-6/5.0 - 2.0*10^-6/5.0) = 3600 V

V_net = 3.6 k

To calculate electric potential at origin, then

V0_net = k*q1/R1 + k*q2/R2

R1 = 3.0 m & R2 = 1.0 m, then

V0_net = 9*10^9*(4.0*10^-6/3.0 - 2.0*10^-6/1.0) = -6000 V

generally it is assumed that electric potential at origin is zero, but in this case it's mentioned that we need to recalibrate potential at origin to zero

Now since ask that we need electric potential at point A with relative to zero at the origin, So if we re-calibrate electric potential to zero at the origin, then we need to add 6000 V everywhere, therefore

Net electric potential at Point A = V_net - V0_net = 3600 - (-6000) = 9600 V = 9.6 kV

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.