Question

Two charged particles attract each other with a force of magnitude F. If the distance between the charges is made 3.5 times as large and the charge on one of the particles is made 4 times as big, what is the ratio of the new F to the old one

Answer 1

Answer:

The ratio between the forces is:

$\frac{F_{new}}{F_{old}}=0.33$

Explanation:

The electrostatic force equation is:

$F_{old}=k\frac{q_{1}q_{2}}{d^{2}}$

Where:

q1 and q2 are the electric charges

d is the distance between them

k is the electrostatic constant

Now, the distance is 3.5 times as large and q1 is 4 times as big, then the new force will be:

$F_{new}=k\frac{4q_{1}q_{2}}{(3.5d)^{2}}$

$F_{new}=\frac{4}{3.5^{2}}k\frac{q_{1}q_{2}}{d^{2}}$

We can rewrite this equation in terms of F(old)

$F_{new}=\frac{4}{3.5^{2}}F_{old}$

Therefore, the ratio between the forces is:

$\frac{F_{new}}{F_{old}}=\frac{4}{3.5^{2}}$

$\frac{F_{new}}{F_{old}}=0.33$

I hope it helps you!