The field just outside a 2.65-cm-radius metal ball is432 N/C and points toward the ball.What charge resides on the ball?

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electreto05 | College Teacher | (Level 1) Assistant Educator

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Given the geometry of the problem, a way to perform the calculation of the charge, it is by applying Gauss's law of electrostatics.

E . da = Q/Ԑ0


E, is the field strength

da, is the vector that characterizes an element of the Gaussian surface.

Q, is the charge, which encloses the surface.

Ԑ0,  is the electric permittivity of vacuum.

To apply this equation, we choose a closed spherical surface adjusted to the outer points of the charged sphere, so as to enclose the entire electric charge. Since the vector (E) points to the inside of the sphere and the vector (da) is directed outwards, the angle between them is 180°. Substituting in the integral, we have:

∫ E da cos 180° = Q/Ԑ0

The field has the same value for all points of the Gaussian surface, so, we just have to integrate over the entire surface area:

- E ∫ da = Q/Ԑ0

- E (4πr^2) = Q/Ԑ0

Q = -[Ԑ0*E*(4πr^2)]

Q = -[(8.854*10-12)(432)(12.56)(7.02*10^-4)]

Q = - 3.37*10^-11 C

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