The electric field at a distance r outside the conductor is :

\frac{1}{4 π ε_{0}} \frac{q_{b}}{r^{2}}

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\frac{1}{4 π ε_{0}} \frac{q_{a} + q_{b}}{r^{2}}

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\frac{1}{4 π ε_{0}} \frac{q_{a}}{r^{2}}

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Solution

The correct option is C $\frac{1}{4 π ε_{0}} \frac{q_{a} + q_{b}}{r^{2}}$
As shown in the figure, total charge on the outer surface of conductor is $Q_{n e t} = q_{a} + q_{b}$
Therefore electric field outside the conductor = $\frac{1}{4 π ϵ_{0}} \frac{q_{a} + q_{b}}{r^{2}}$

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