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Electric Charges and Fields — Physics STD 12 Science — Question

Gujarat 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.
  • A
    Both A and R are true, and R is the correct explanation of A.
  • ✓
    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: B.
Both A and R are true, but R is not the correct explanation of A.
  1. Both A and R are true, but R is not the correct explanation of A.
Explanation:
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 $\mathrm{r}_1$ and $\mathrm{r}_2$ be two different radii) Then the ratio of electric field intensities near the surface of spherical conductors,
$\frac{\text{E}_1}{\text{E}_2}=\frac{\text{q}_1}{4\pi\in_0\text{r}^2_1}\times\frac{4\pi\in_0\text{r}^2_2}{\text{q}_2}$
$=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{r}^2_2}{\text{r}^2_1}=\frac{\text{q}_1}{\text{q}_2}\times\frac{\text{q}_2}{\text{q}_1}=1$
i.e. $\mathrm{E}_1=\mathrm{E}_2$.

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