Question

solve this asap
1.if both the magnitude of two charges and distance between them is doubled then new electrostatic force will be..
a. F
b. 2F
c. zero
d. none of these
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Answer 1

Answer:

B..... 2f is the correct answer

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Answer 2

Answer:

The answer will be F (OPTION A)

Explanation:

We Can solve this problem with the help of COULOMB'S LAW. The law states, "the amount of electric force between two stationary charged particles is directly proportional to the product of charges and inversely proportional to the square of the distance between them."

Coulomb's law formula

$F=k\frac { |Q_1|\times |Q_2|}{r^2}$

Where:

F = Coulombic force expressed by the charges in Newtons (N).

Q1 = first point charge expressed in Coulombs (C).

Q2 = the second point charge (C).

r =   the distance between two point charges expressed in meters (m).

k  = electrostatic constant whose value is approximately 8.988 × 10 9 N · m 2 · C -2

So let's suppose there are two charges, Q1 and Q2, and the distance between them is r. So the Coloumbic force between them is -

                             $F=k\frac { |Q_1|\times |Q_2|}{r^2}$

Now according to the question, if Q1 becomes 2Q1 and Q2 becomes 2Q2 and r becomes 2r. So the new force $(F_1)$ between them becomes-

   $F_1=k\frac { |2Q_1|\times |2Q_2|}{(2r)^2}\\F_1=k2 \times2 \frac { |Q_1|\times |Q_2|}{4r^2}\\ F_1=k\frac { |Q_1|\times |Q_2|}{(r)^2}=F$

HENCE, THE CORRECT ANSWER IS  F (OPTION A).