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jekas [21]
3 years ago
13

A solid sphere of radius R carries a fixed, uniformly distributed charge q. Obtain an expression for the magnitude of the electr

ic field created by the sphere at a point P outside the sphere.
Physics
1 answer:
NISA [10]3 years ago
4 0

Answer:

The electric field outside the sphere will be \dfrac{qr}{4\pi\epsilon_{0}R^3}.

Explanation:

Given that,

Radius of solid sphere = R

Charge = q

According to figure,

Suppose r is the distance between the point P and center of sphere.

If \rho be the volume charge density,

Then, the charge will be,

q=\rho\times\dfrac{4}{3}\pi R^3.....(I)

Consider a Gaussian surface of radius r.

We need to calculate the electric field outside the sphere

Using formula of electric field

\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}

E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}

Put the value from equation (I)

E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}

E=\dfrac{qr}{4\pi\epsilon_{0}R^3}

Hence, The electric field outside the sphere will be \dfrac{qr}{4\pi\epsilon_{0}R^3}.

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