If the force between two charged objects is to be left unchanged, even though the charge on one of the objects is halved, keeping the other the same, the original distance of separation 'd' should be changed to:

d / \sqrt{2}

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d / 2

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\sqrt{2} d

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2 d

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Solution

The correct option is A $d / \sqrt{2}$
$F α \frac{q_{1} q_{2}}{d^{2}}$
Given $q_{1} = q_{1} / 2; q_{2} = q_{2}$
$∴ F^{'} = k \frac{q_{1} q_{2}}{2 d^{' 2}}$
$\frac{k q_{1} q_{1}}{2 (d^{1})^{2}} = \frac{k q_{1} q_{2}}{d^{2}}$
$d^{2} = 2 (d^{1})^{2}$
$\frac{d}{\sqrt{2}} = d^{'}$

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Similar questions

If the force between two charged objects is to be left unchanged even though the charge on one of the object is halved keeping the other same, the original distance (d) of separation should be changed to :

Q. If the force between two charged objects separated by a distance d is to be left unchanged even though the charge on one of the objects is halved, keeping other the same, the new distance of separation becomes :

The gravitational force of attraction between two objects is x. Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of gravitational force between them will become :

(a) $x / 4$
(b) $x / 2$
(c) $2 x$
(d) $4 x$

$Q_{1}, Q_{2}$ charges are separated by distance $‘ d^{'}$ . $F$ is the force between them. Now the value of $Q_{2}$ is halved. To have the same force between the charges, the distance of separation should be :

Q. Two point charges separated by a distance d repel each other with a force of

10 N

. If the separation between them becomes

d / 2

, the force of repulsion will be ______.

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If the force between two charged objects is to be left unchanged, even though the charge on one of the objects is halved, keeping the other the same, the original distance of separation 'd' should be changed to:

The correct option is A d/√2Fαq1q2d2Given q1=q1/2;q2=q2∴F′=kq1q22d′2kq1q12(d1)2=kq1q2d2d2=2(d1)2d√2=d′

The correct option is A $d / \sqrt{2}$
$F α \frac{q_{1} q_{2}}{d^{2}}$
Given $q_{1} = q_{1} / 2; q_{2} = q_{2}$
$∴ F^{'} = k \frac{q_{1} q_{2}}{2 d^{' 2}}$
$\frac{k q_{1} q_{1}}{2 (d^{1})^{2}} = \frac{k q_{1} q_{2}}{d^{2}}$
$d^{2} = 2 (d^{1})^{2}$
$\frac{d}{\sqrt{2}} = d^{'}$