The field at P outside the sphere due to induced charge on the sphere is

\frac{K q_{1}}{r_{1}^{3}}

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K q (\frac{_{1}}{r_{1}^{3}} + \frac{_{2}}{r_{2}^{3}})

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K q (\frac{_{1}}{r_{1}^{3}} - \frac{_{2}}{r_{2}^{3}})

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K q (- \frac{_{1}}{r_{1}^{3}} + \frac{_{2}}{r_{2}^{3}})

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Solution

The correct option is A $\frac{K q_{1}}{r_{1}^{3}}$
q inside cavity will induce -q on the inner walls of cavity and +q at the outer walls of cavity.
Thus, at point P outside the spheere the electric field wil be simply
$\to E = \frac{1}{4 π ϵ_{0}} \frac{q}{{r_{1}}^{3}}_{1}$

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The field at P outside the sphere due to induced charge on the sphere is

The correct option is A Kq→r1r31q inside cavity will induce -q on the inner walls of cavity and +q at the outer walls of cavity.Thus, at point P outside the spheere the electric field wil be simply→E=14πϵ0qr13→r1

The correct option is A $\frac{K q_{1}}{r_{1}^{3}}$
q inside cavity will induce -q on the inner walls of cavity and +q at the outer walls of cavity.
Thus, at point P outside the spheere the electric field wil be simply
$\to E = \frac{1}{4 π ϵ_{0}} \frac{q}{{r_{1}}^{3}}_{1}$