AP EAMCET · PHYSICS · Electrostatics
In space the electric potential varies as \(V=20|\vec{r}|\) volt, where \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) is the position vector. Then electric field in \(\left(\mathrm{N} \mathrm{C}^{-1}\right)\) at the point \((4 \mathrm{~m}, 3 \mathrm{~m},-5 \mathrm{~m})\) is
- A \(-\sqrt{2}(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-10 \hat{\mathrm{k}})\)
- B \(-\sqrt{2}(8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-10 \hat{\mathrm{k}})\)
- C \(-(8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-10 \hat{\mathrm{k}})\)
- D \(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}\)
Answer & Solution
Correct Answer
(B) \(-\sqrt{2}(8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-10 \hat{\mathrm{k}})\)
Step-by-step Solution
Detailed explanation
\(\vec{E} = -\nabla V\) \(V = 20|\vec{r}| = 20\sqrt{x^2+y^2+z^2}\) \(\nabla V = \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k}\)…
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