12th Class Physics Electric Charges and Fields Question Bank Case Based (MCQs) - Electric Charges and Fields

  • question_answer
    Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason:
    (R). Mark the correct choice as:
    Assertion (A): The surface charge densities of two spherical conductors of different radii are equal. Then the electric field intensities near their surface are also equal.
    Reason (R): Surface charge density is equal to charge per unit area.

    A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)

    B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)

    C) Assertion (A) is true but Reason (R) is false            

    D) Assertion (A) is false and Reason (R) is also false

    Correct Answer: B

    Solution :

    (b) As \[{{\sigma }_{1}}={{\sigma }_{2}}\]                                    (Given) \[\frac{{{q}_{1}}}{4\pi r_{1}^{2}}=\frac{{{q}_{2}}}{4\pi r_{2}^{2}}\] or \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{r_{1}^{2}}{r_{2}^{2}}\] Then ratio of electric field intensities, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{q}_{1}}}{4\pi {{\varepsilon }_{0}}r_{1}^{2}}\times \frac{4\pi {{\varepsilon }_{0}}r_{2}^{2}}{{{q}_{2}}}\] \[=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{r_{2}^{2}}{r_{1}^{2}}=\frac{{{q}_{1}}}{{{q}_{2}}}\times \frac{{{q}_{2}}}{{{q}_{1}}}=1\] i.e., \[{{E}_{1}}={{E}_{2}}\]


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