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  • Need clarity, kindly explain! Two equal spheres are identically charged with q units of electricity separately. When they are placed at a distance 3R from centre-to-centre where R is the radius of either sphere the force of repulsion between them is

Two equal spheres are identically charged with q units of electricity separately. When they are placed at a distance 3R from centre-to-centre where R is the radius of either sphere the force of repulsion between them is

  • Option 1)

    \frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{R^{2}}\; \;

  • Option 2)

    \; \frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{9R^{2}}\; \;

  • Option 3)

    \; \frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{4R^{2}}

  • Option 4)

    None of these

 

Answers (2)

best_answer

As we learned

By electrostatic induction -

When charged body brought near an uncharged body, one side of neutral body becomes oppositely charged while other becomes same charged.

- wherein

 

 Generally students give the answer \frac{1}{4\pi \epsilon _{0}}\frac{q^{2}}{(3R)^{2}}   but it is not true. Since the charges are not uniformly distributed, they cannot be treated as point charges and so we cannot apply coulombs law which is a law for point charges. The actual distribution is shown in the figure above.

 


Option 1)

\frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{R^{2}}\; \;

Option 2)

\; \frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{9R^{2}}\; \;

Option 3)

\; \frac{1}{4\pi \varepsilon _{0}}.\frac{q^{2}}{4R^{2}}

Option 4)

None of these

Posted by

Aadil

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

4)NONE OF THESE 

REASON:

1) THE BODIES ARE VERY LARGE

2) THEIR DISTANCE OF SEPARATION  IS COMPARABLE TO THEIR SIZE

CANNOT USE COULOMBS LAW, BECAUSE IT IS APPLIED ONLY FOR POINT CHARGES.

3) SO THE CHARGES WOULD REPEL AND DISTRIBUTION OF CHARGE WILL BE LESS NEAR THE TWO AND MORE AT THE FARTHER ENDS OF THE SPHERES .

 

Posted by

MOUSUMI MUKHERJEE

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