Question

Two point charges +2c and +4c repel to each other with a force of 8N. If the charge of -1c is given to each of these charges, the forces will be.... (k/r2=1)

Answer 1

Explanation:

The new force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is \bold{4 \times 10^{-6} N}4×10−6N .

Solution:

The old charges are +2μC and +6μC. If an additional charge of -4μC is added in that then the new charges will be -2μC and 2μC respectively.

The force between the two old point charges is 12N. As they both are positive charges, they repel each other. So the distance between the two charges can be found from coulomb’s law.

F=k \frac{q q^{\prime}}{r^{2}}F=kr2qq′

12=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times 6 \times 10^{-6}}{r^{2}}12=9×109×r22×10−6×6×10

r^{2}=\frac{9 \times 2 \times 6 \times 10^{3}}{12}=9 \times 10^{3}r2=129×2×6×103=9×103

The distance found can be used in the force between the new charges -2μC and 2μC.

F=9 \times 10^{9} \times \frac{2 \times 10^{-6} \times-2 \times 10^{-6}}{9 \times 10^{3}}F=9×109×9×1032×10−6×−2×10−6

F=-\frac{9 \times 2 \times 2 \times 10^{9-6-6-3}}{9}=-4 \times 10^{-6} NF=−99×2×2×109−6−6−3=−4×10−6N

So, the new force is attractive in nature as the charges are in opposite sign and the magnitude of the new force is \bold{4 \times 10^{-6} \mathrm{N}.}4×10−6N.

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Answer 2

Answer:

New force is 3N.

Explanation:

Coulomb's law states that the force of attraction or repulsion between two charged bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. It operates on the segment connecting the two charges regarded as point charges.

When two objects behave as point charges, the force between them can be accurately described by the Coulomb's law equation. When interacting with other charged objects, a charged conducting sphere behaves as though its whole charge were concentrated in its center. Although the charge is evenly distributed around the sphere's surface, the center of charge can be thought of as the sphere's center. The excess charge at the sphere's center causes it to function as a point charge. The distance d in the equation is the separation between the center's of charge for both objects because Coulomb's law only applies to point charges (not the distance between their nearest surfaces).

$F=\frac{KQ_{1} Q_{2} }{d^{2} }$.........(1)

$Q_{1} ,Q_{2}$ are the charges

d is the distance

Given:

Charge $Q_{1}=+2C$

Charge, $Q_{2} =4C$

Condition given charge of -1c is given to each of these charges, therefore

$Q_{1} =2-1C=2C\\Q_{2} =4-1C=3C$

$\frac{k}{r^{2} } =1$

Substituting the above values in equation (1), we get-

$F=\frac{1\times3}{1} \\F=3N$

Hence, new force is 3N.

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