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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

There is a so...

Question

There is a solid sphere of radius $R$ having uniformly distributed charge throughout it. What is the relation between electric field $E$ and distance $r$ from the centre $(r < R)$ ?

E \propto r^{- 2}

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E \propto r^{- 1}

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E \propto r

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E \propto r^{2}

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Solution

The correct option is B $E \propto r^{- 2}$
There is a solid sphere radius $= R$
uniformly distributed charge through out it $r < R$
The relation will be $E α \frac{1}{r^{2}}$ because the formula occurs :-
$E = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}$

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Similar questions

Q. Charge

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Q. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E

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The electric potential inside a charged sphere varies with the distance 'r' from its centre as V = a - br $^{3}$ , where 'a' and 'b' are positive constant. Then select the correct statement(s) from the following.

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There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r<R)?

The correct option is B E∝r−2There is a solid sphere radius =Runiformly distributed charge through out it r<RThe relation will be Eα1r2 because the formula occurs :-E=14πϵ0qr2

There is a solid sphere of radius $R$ having uniformly distributed charge throughout it. What is the relation between electric field $E$ and distance $r$ from the centre $(r < R)$ ?

The correct option is B $E \propto r^{- 2}$
There is a solid sphere radius $= R$
uniformly distributed charge through out it $r < R$
The relation will be $E α \frac{1}{r^{2}}$ because the formula occurs :-
$E = \frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}$