Electric potential in an electric field is given as V = K r , (K being constant), if…
Physics · JEE Main · NTA Exams — Electrostatics
Electric potential in an electric field is given as \(\mathrm{V}=\frac{\mathrm{K}}{\mathrm{r}},\) (K being constant), if position vector \(\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}},\) then electric field will be
- \((2 \hat{i}+3 \hat{j}+6 \hat{k}) \frac{K}{243}\)
- \((2 \hat{i}+3 \hat{j}+6 \hat{k}) \frac{K}{343}\)
- \(\frac{\mathrm{K}}{243}(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})\)
- \(\frac{\mathrm{K}}{343}(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\)
Answer
(B) (2 i +3 j +6 k ) K 343
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