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Two point charge \(\mathrm{Q}\) and \(-3 \mathrm{Q}\) are placed at some distance apart. If the electric field at the location of \(Q\) is \(\vec{E}\), the field at the location of \(-3 Q\) is:
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Verified Answer
The correct answer is:
\(+\frac{\vec{E}}{3}\)
Given Electric field at \(\mathrm{A}=\overrightarrow{\mathrm{E}}\)
\(\overrightarrow{\mathrm{E}}:\) Magnitude & direction
Magnitude : \(\mathrm{E}\)
direction : \((\mathrm{x}-\) axis)
Now, \(\mathrm{E}=\frac{\mathrm{K}(3 \mathrm{Q})}{\mathrm{r}^2}\{\) Electric field due to charge at \(\mathrm{B}\}\)
or \(\frac{\mathrm{KQ}}{\mathrm{r}^2}=\mathrm{E}^1\left\{\mathrm{E}^1\right.\) : Electric field magnitude at \(\mathrm{B}\) due to \(\left.\mathrm{A}\right\}\)
\(\therefore \mathrm{E}^1=\frac{\mathrm{E}}{3}\)
for direction at \(\mathrm{E}^1, \therefore \mathrm{Q}\) is a + ve charge particle & electric field will be along \(+\mathrm{x}-\) axis,
\(\therefore \overrightarrow{\mathrm{E}}=+\frac{\overrightarrow{\mathrm{E}}}{3}\) {or in direction of \(\overrightarrow{\mathrm{E}}\) itself}
\(\overrightarrow{\mathrm{E}}:\) Magnitude & direction
Magnitude : \(\mathrm{E}\)
direction : \((\mathrm{x}-\) axis)
Now, \(\mathrm{E}=\frac{\mathrm{K}(3 \mathrm{Q})}{\mathrm{r}^2}\{\) Electric field due to charge at \(\mathrm{B}\}\)
or \(\frac{\mathrm{KQ}}{\mathrm{r}^2}=\mathrm{E}^1\left\{\mathrm{E}^1\right.\) : Electric field magnitude at \(\mathrm{B}\) due to \(\left.\mathrm{A}\right\}\)
\(\therefore \mathrm{E}^1=\frac{\mathrm{E}}{3}\)
for direction at \(\mathrm{E}^1, \therefore \mathrm{Q}\) is a + ve charge particle & electric field will be along \(+\mathrm{x}-\) axis,
\(\therefore \overrightarrow{\mathrm{E}}=+\frac{\overrightarrow{\mathrm{E}}}{3}\) {or in direction of \(\overrightarrow{\mathrm{E}}\) itself}
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