A 6.00 nC is 2.00 m from a 3.00 nC charge. Find the magnitude of the electric field at a point midway between the two charges? Which way does the electric field point, towards the positive or the negative charge?

E_total = 26.97 N/C

Electric field points towards the positive charge

Explanation:

We are given;

Charge 1; q1 = 6 nC = 6 × 10^(-9) C

Charge 2; q2 = 3 nC = 3 × 10^(-9) C

Distance between both charges; R_o = 2 m

Since we want to find electric field midway, the distance midway is r = 2/2 = 1 m

Using coulumbs law;

E = kq/r²

Where k is a constant with a value of 8.99 × 10^(9) N.m/C²

Thus;

E1 = kq1/r²

E1 = (8.99 × 10^(9) × 6 × 10^(-9))/1²

E1 = 53.94 N/C

Similarly;

E2 = kq2/r²

E2 = (8.99 × 10^(9) × 3 × 10^(-9))/1²

E2 = 26.97 N/C

Since both electric fields are positive, it means that they are both moving towards the midpoint of the distance between both charges.

This implies they will have opposite directions.

Thus, total electric field at the midway point is;

E_total = E1 - E2

E_total = 53.94 - 26.97

E_total = 26.97 N/C

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.

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