KCET · Physics · Electrostatics
The potential of the electric field produced by point charge at any point \((x, y, z)\) is given by \(V=3 x^{2}+5\), where \(x, y\) are in metres and \(V\) is in volts. The intensity of the electric field at \((-2,1,0)\) is
- A \(+17 \mathrm{Vm}^{-1}\)
- B \(-17 \mathrm{Vm}^{-1}\)
- C \(+12 \mathrm{Vm}^{-1}\)
- D \(-12 \mathrm{Vm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(-12 \mathrm{Vm}^{-1}\)
Step-by-step Solution
Detailed explanation
Potential \((V)=3 x^{2}+5\) Intensity of the electric field \(=\frac{d V}{d x}=6 x\)
\(\therefore E \text { at } x =-2\)
\(=6(-2)=-12 \mathrm{~V} / \mathrm{m}\)
\(\therefore E \text { at } x =-2\)
\(=6(-2)=-12 \mathrm{~V} / \mathrm{m}\)
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