MHT CET · Physics · Electrostatics
The electric potential ' \(V\) ' is given as a function of distance ' \(x\) ' (metre) by \(V=\left(4 x^2+8 x-3\right) V\). The value of electric field at \(x=0.5 \mathrm{~m}\), in \(\mathrm{V} / \mathrm{m}\) is
- A \(-16\)
- B \(-12\)
- C \(0\)
- D \(+12\)
Answer & Solution
Correct Answer
(B) \(-12\)
Step-by-step Solution
Detailed explanation
\(\frac{dV}{dx} = \frac{d}{dx}\left(4 x^2+8 x-3\right) = 8x + 8\) \(E = -\frac{dV}{dx} = -(8x + 8)\)
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