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Electric Charges and Fields — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields1 Mark
MCQ
For two statements are given$-$one labelled Assertion $(A)$ and the other labelled Reason $(R).$ Select the correct answer to these questions from the codes $(a), (b), (c)$ and $(d)$ as given below.
Assertion $(A):$ The surface densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal.
Reason $(R):$ Surface density is equal to charge per unit area.
  • ✓
    Both $A$ and $R$ are true, and $R$ is the correct explanation of $A.$
  • B
    Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A.$
  • C
    $A$ is true but $R$ is false.
  • D
    $A$ is false and $R$ is also false.
✓

Answer

Correct option: A.
Both $A$ and $R$ are true, and $R$ is the correct explanation of $A.$
As$, \sigma_1=\sigma_2 ($Given$)$
$\therefore\frac{\text{q}_1}{4\pi\text{r}_1^2}=\frac{\text{q}_2}{4\pi\text{r}_2^2},$
Or $\frac{\text{q}_1}{\text{q}_2}=\frac{\text{r}_1^2}{\text{r}^2_2}$
$[$ Let $r_1$ and $r_2$ be two different radii $]$
Then the ratio of electric field intensities near the surface of spherical conductors,
$\frac{\text{E}_1}{\text{E}_2}=\frac{\text{q}_1}{4\pi\epsilon_0\text{r}^2_1}\times\frac{4\pi\epsilon_0\text{r}^2_2}{\text{q}_2}$
$=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{r}^2_2}{\text{r}^2_1}$
$=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{q}_2}{\text{q}_1}=1$
i.e. $E_1 = E_2.$

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