The electric field intensity at $P$ and $Q,$ in the shown arrangement:

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Solution

Drawing Gaussian surface (spherical) at point $P$ of radius $r$ and at point $Q$ of radius $2 r$
Now Applying Gauss's law
For point $P$
$\oint_{p} . \to d s = \frac{V_{i n}}{ϵ_{0}} \Rightarrow E_{p} \int d s = \frac{V}{ϵ_{0}} \Rightarrow E_{p} .4 π r^{2} = \frac{V}{ϵ_{0}} \Rightarrow E_{p} = \frac{V}{4 π r^{2} ϵ_{0}}$
for point $Q$
$\oint_{Q} . \to d s = \frac{V_{i n}}{ϵ_{0}}$
$E_{Q} \int d s = \frac{V + 3 V}{ϵ_{0}} \Rightarrow E_{p} .4 π {(2 r)}^{2} = \frac{4 V}{ϵ_{0}} \Rightarrow E_{p} = \frac{V}{4 π r^{2}}$

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The electric field intensity at P and Q, in the shown arrangement:

Drawing Gaussian surface (spherical) at point P of radius r and at point Q of radius 2rNow Applying Gauss's lawFor point P∮→Ep.→ds=Vinϵ0⇒Ep∫ds=Vϵ0⇒Ep.4πr2=Vϵ0⇒Ep=V4πr2ϵ0for point Q∮→EQ.→ds=Vinϵ0EQ∫ds=V+3Vϵ0⇒Ep.4π(2r)2=4Vϵ0⇒Ep=V4πr2

The electric field intensity at $P$ and $Q,$ in the shown arrangement: