Question

A charge of +3.0 µC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25µC from infinity to the center of the circle?

Answer 1

Answer:

3.37x10^6J

Explanation:

The external energy is given as KQ1Q2/r

=(9*10^9)*(3*10^-3)*(25*10^-3)/(0.2)

=3.375 *10^6 J

Answer 2

Answer:

The energy required is  $E= 3.375*10^6J$

Explanation:

From the question we are told that

  The charge is $q_1 = 3.0 \mu C = 3.0*10^{-6}C$

  The radius is $r = 20cm = \frac{20}{100} = 0.2m$

   The second charge is $q_2 = 25 \mu C = 25*10^{-6}C$

Generally the energy required is mathematically represented as

                          $E = \frac{k q_1 q_2 }{R}$

Where k is the coulomb constant with a value of  $k = 9*10^9kg m^3 s^{-4} A^{-2}$

       Substituting values we have

                         $E = \frac{9*10^9 * 3*10^{-3} * 25 *10^{-6}}{0.2}$

                            $E= 3.375*10^6J$