Question

Two charged particles having equal charge of 3×10–5C each are brought from infinity to a separation of 30cm the increase in electrostatic potential energy during process is …. a) 90J b) 27J c) 3J d) 1J​

Answer 1

Answer:

The magnitude of the two charges is same q1=q2=2×10−5 C

Each are brought from infinity to 10cm a part 

So, the distance covered by the charge is d=10×10−2m

So work done=negative of work done.(Potential E)

P.E=∫∞10F×dx

P.E=k×rq1q2

=10×10−29×109×4×10−10=36J.

Step-by-step explanation:

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Answer 2

Concept:

Electric potential energy is possessed by an object by the virtue of two elements, those being, the charge possessed by an object itself and the relative position of an object with respect to other electrically charged objects. The magnitude of electric potential depends on the amount of work done in moving the object from one point to another against the electric field.

When an object is moved against the electric field it gains some amount of energy which is defined as the electric potential energy. For any charge, the electric potential is obtained by dividing the potential energy by the quantity of charge.

U=kq₁q₂/r

Given:

Two charged particles having equal charge of 3×10–5C each are brought from infinity to a separation of 30cm

Find:

the increase in electrostatic potential energy during process is …. a) 90J b) 27J c) 3J d) 1J​

Solution:

The magnitude of the two charges is same q1=q2=3×10⁻⁵ C

Each are brought from infinity to 30cm apart

So, the distance covered by the charge is d=30×10⁻²m

So,Work done= negative of change in potential energy

W=-ΔU

P.E=kq₁q₂/r

     =(9 x 10⁹ x 3×10⁻⁵ x 3×10⁻⁵ )/0.3

     =27J

Therefore , change in potential energy is 27J

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