Question

A +25µC point charge is placed 6.0cm from an identical +25µC point charge. How much work would be required by an external force to move a +0.18µC test charge from a point midway between them to a point 1.0cm closer to either charge?

Answer 1

Answer:

Explanation:

Required work done = charge moved x potential difference at two points

Potential at mid point

= 2 x k q / r

= 2 x 9 x 10⁹ x 25 x 10⁻⁶ / 3 x 10⁻²

= 150 x 10⁵ V

Potential at the second point

=  9 x 10⁹ x 25 x 10⁻⁶ / 4 x 10⁻²  + 9 x 10⁹ x 25 x 10⁻⁶ / 2 x 10⁻²

= 56.25 x 10⁵ + 112.5 x 10⁵

= 168.75 x 10⁵ V

Potential difference

= (168.75 - 150 ) x 10⁵ V

= 18.75 x 10⁵

Work done

= .18 x 10⁻⁶ x 18.75 x 10⁵ J

= 3.375 x 10⁻¹ J

= .3375 J

Answer 2

The work done needed by an external force is .3375 J

calculation of work required:

Required work done = charge moved × potential difference at two points

Here

Potential at midpoint

= 2 x 9 x 10⁹ x 25 x 10⁻⁶ / 3 x 10⁻²

= 150 x 10⁵ V

Potential at the second point

=  9 x 10⁹ x 25 x 10⁻⁶ / 4 x 10⁻²  + 9 x 10⁹ x 25 x 10⁻⁶ / 2 x 10⁻²

= 56.25 x 10⁵ + 112.5 x 10⁵

= 168.75 x 10⁵ V

Potential difference

= (168.75 - 150 ) x 10⁵ V

= 18.75 x 10⁵

So, Work done

= .18 x 10⁻⁶ x 18.75 x 10⁵ J

= 3.375 x 10⁻¹ J

= .3375 J

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