Question

Two point charged particles are 4.41 cm apart. They are moved and placed in a new position. The force between them is found have tripled. How far apart are they now (in cm)?​

Answer 1

Answer:

Two point charged particles are 4.41 c m 4.41\,{\rm cm} 4.41cm apart. They are moved and placed in a new position. The force between them is found to have tripled. ... Solution: initial distance is r 1 = 4.41 c m r_1=4.41\,{\rm cm} r1=4.41cm.

may it will help you dear....

Answer 2

Answer:

The charges are 2.546cm apart

Explanation:

Given that,

Two point charges are initially 4.41cm apart

The Electrostatic force between two charges is given by,

$F = \frac{1}{4\pi ε_{0} } \frac{ q_{1} q_{2} }{r {}^{2} }$

where,

q1 and q2 are point charges

r=Distance between the charges

From the relation,we can interpret that the Electrostatic force is inversely proportional to square of distance between two charges

Therefore,

$F∝ \frac{1}{r {}^{2} } \\ F_{1} r_{1} {}^{2} = F_{2} r_{2} {}^{2}$

Given that,the force between charges is tripled.Hence,

$F(4.41) {}^{2} = 3F r_{2} {}^{2} \\ = = > r_{2} = 2.546cm$

Therefore,the charges are 2.546cm apart

#SPJ2