KCET · Physics · Electrostatics
The electric potential at any point \(x, y, z\) in metres is given by \(\mathrm{V}=3 \mathrm{x}^{2}\). The electric field at a point \((2,0,1)\) is
- A \(12 \mathrm{Vm}^{-1}\)
- B \(-6 \mathrm{Vm}^{-1}\)
- C \(6 \mathrm{Vm}^{-1}\)
- D \(-12 \mathrm{Vm}^{-1}\)
Answer & Solution
Correct Answer
(D) \(-12 \mathrm{Vm}^{-1}\)
Step-by-step Solution
Detailed explanation
Here, electric potential
\(V=3 x^{2}\)
Electric field,
\(\mathrm{E}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} =-\frac{\partial}{\partial \mathrm{x}}\left(3 \mathrm{x}^{2}\right)=-6 \mathrm{x} \)
\(\therefore \mathrm{E}_{(2,0,1)} =-12 \mathrm{Vm}^{-1}\)
\(V=3 x^{2}\)
Electric field,
\(\mathrm{E}=-\frac{\partial \mathrm{V}}{\partial \mathrm{x}} =-\frac{\partial}{\partial \mathrm{x}}\left(3 \mathrm{x}^{2}\right)=-6 \mathrm{x} \)
\(\therefore \mathrm{E}_{(2,0,1)} =-12 \mathrm{Vm}^{-1}\)
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