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Standard XII

Physics

Superposition of Electric Fields

The electric ...

Question

The electric field at the surface of a charged spherical conductor is $10 K V ? m$ . The electric field at a distance equal to the diameter from its centre will be-

2.5 V / m

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2.5 K V / m

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5.0 K V / m

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5.0 V / m

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Solution

The correct option is C $5.0 K V / m$
$V = 10 K V$
radius $= r$
diameter $= 2 r$
hence $E = \frac{d v}{d r} = \frac{10}{2} = 5.0 K V / m$

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The electric field at the surface of a charged spherical conductor is 10KV?m. The electric field at a distance equal to the diameter from its centre will be-

The correct option is C 5.0KV/mV=10KVradius =rdiameter =2rhence E=dvdr=102=5.0KV/m

The electric field at the surface of a charged spherical conductor is $10 K V ? m$ . The electric field at a distance equal to the diameter from its centre will be-

The correct option is C $5.0 K V / m$
$V = 10 K V$
radius $= r$
diameter $= 2 r$
hence $E = \frac{d v}{d r} = \frac{10}{2} = 5.0 K V / m$