Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. the electric field at a point p, which is 0.50 m from the center of the sphere, is 15,000 n/c and is directed radially outward. what is the maximum magnitude of the electric field due to this sphere?

Explanation:

Given that,

The radius of sphere, r = 0.3 m

Distance from the center of the sphere to the point P, x = 0.5 m

Electric field at point P, (radially outward)

The maximum electric field is at the surface of the sphere. We know that the electric field is inversely proportional to the distance. So,

So, the magnitude of the electric field due to this sphere is 41666.66 N/C. Hence, this is the required solution.

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initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft