The flux of a 33 `microC charge spread out uniformly over concentric sphere. On a sphere of radius 3.9 meters, what is the total flux and what is the flux per unit area? What is the strength of the electric field on this sphere?
Answer the same question for a sphere of radius r, with a charge q at its center. Use your answers to determine the effect on the electric field of doubling the distance from the charge, of doubling the charge without changing the distance, and of doubling both distance and charge.
- flux / area = ( 3732 x 10^3 N m^2 / C) / ( 191.1 m^2) = 19.52 * 10^3 N / C.
If we double the distance r from a charge, the area 4`pi r^2 of the sphere over which the flux is spread quadruples. The flux will therefore be divided by four times the area, and will be only as 1/4 great as before.
If the charge is doubled, the flux will be doubled. If the distance is not changed, the field strength will be equal to a doubled flux divided by the same area. Thus the field strength will be doubled.
Doubling the distance quadruples the area while doubling the charge doubles the flux. When we divide a doubled flux by a quadrupled area we get half the former electric field.
The flux of a charge Q is 4 `pi k Q. When spread out over a sphere of radius r, the flux density is 4 `pi k Q / (area of the sphere) = 4 `pi k Q / (4`pi r^2) = k Q / r^2. This is the electric field strength.
At distance 2r, the field is similarly found to be k Q / (2r)^2 = (1/4) k Q / r^2, or 1/4 as great as at distance r.
If charge is doubled the field will be k (2Q) / r^2, twice as great as for the original charge at this distance.
If distance and charge are both doubled the field becomes k (2Q) / (2r)^2 = (1/2) k Q / r^2, half as great as for the original charge and distance.
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