Answer to Question #137225 in Mechanics | Relativity for Sneha

Question

Explain electrostatic potential energy.

Answer 1

Charged bodies attract or repel each other according to Kulon law $F = k\times\frac {q _ 1\times q_2}{r^2}$ , when the charged bodies move, the forces acting between them work, for small movement $∆l$ the formula of work can be written as follows: $ΔA = F\times Δl\times\cos α =$ $E\times q\times Δl\times\times α =$ $E _ q\times Δl\times ∆A=$ $F\times ∆l\times\cos α$ $= E\times q\times ∆l\times\cos α =$ $E _ l\times q\times ∆l$

Where $E$ is the electrostatic field strength, $q$ charge, $Δl$ is the moving from point (1) to point (2). It is known from mechanics that a system capable of doing work due to the interaction of bodies with each other has potential energy. So, the system of charged bodies has a potential energy called electrostatic or electrical. Since the operation of the electrostatic force does not depend on the shape of the trajectory of its application point, the force is conservative, and its work is equal to the change in potential energy taken with the opposite sign: $A = - (W_{p2} - W_{p1}) = -ΔW_p$ . Comparing the previously obtained expression of work $A$ with the general definition of potential energy, we see that $ΔW_p = W_{p2} - W_{p1} = -q\times E\times d$ . We believe that at point (2) the potential energy is zero. Then the potential charge energy in a homogeneous electrostatic field is: $W_p = q\times E\times d$ .