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Q12PE

Expert-verifiedFound in: Page 664

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**Two point charges exert a 5.00 N force on each other. What will the force become if the distance between them is increased by a factor of three?**

If the separation is increased to three times of the original, force will become 0.556N .

- Value of point charges is 5.00 N

Coulomb stated that when two-point charges are separated by some distance, they experience some force of attraction or repulsion. This force of attraction or repulsion is known as electrostatic force.

Initially, when the charges *q* and Q are separated by some distance *r* the electrostatic force between them is,

$F=\frac{KqQ}{{r}^{2}}..........(1.1)$

Here, *F* is the electrostatic force (*F* = 5.00 N) , and *K* is the electrostatic force constant.

When the separation between the charges increased by a factor of three $\left({r}^{\mathbf{\text{'}}}=3r\right)$. The new electrostatic force (F^{'}) between the point charges is,

${F}^{\text{'}}=\frac{KqQ}{{r}^{\text{'}2}}\phantom{\rule{0ex}{0ex}}=\frac{KqQ}{{\left(3r\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{KqQ}{9{r}^{2}}$

Using equation (1.1),

${F}^{\text{'}}=\frac{F}{9}$

Substituting 5.00 N for *F*,

${F}^{\text{'}}=\frac{5.00N}{9}\phantom{\rule{0ex}{0ex}}=0.556N$

Hence, when the separation becomes three times of the original, the new force will be 0.556 N.

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