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Electrostatic Potential Energy - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Electrostatic Potential energy is considered one of the most asked concept.

  • 31 Questions around this concept.

Solve by difficulty

A charged particle q is shot towards another charged particle Q which is fixed, with a speed \nu. It approaches Q up to the closest distance r and then returns. If q were given a speed 2\nu, the closest distances of approach would be

A charged ring has mass mand radius R. A point charge qand mass M is placed at its centre. If total potential energy of the system is zero, then find charge on the ring.

An alpha particle is accelerated through a potential difference of \mathrm{10^6 volt }. Its Kinetic energy will be -

An electromagnetic wave travels along z-axis. which of the following pair of space and time raring fields would generate such a wave?

The potential energy of a system of two particles is given by v=-\frac{m}{r^2}+\frac{n}{r^8}, where r is distance between the particle and m and n are positive constant. If the system is in equilibrium at r=r_0 then r_0 in terms of m and n -

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Concepts Covered - 1

Electrostatic Potential energy

Electrostatic Potential energy -

It is the amount of work done by external forces in bringing a body from  \infty  to a given point against electric force.

or It is defined as negative work done by the electric force in bringing a body from  \infty  to that point.

  • It is Scalar quantity

  • SI Unit: Joule

  • Dimension : \left[ ML^{2}T^{-2}\right ]

Electric Potential energy at a point

If the point charge Q is producing the electric field

Then electric force on test charge q at a distance r from Q is given by F=\frac{KQq}{r^2}

   And the amount of work done by the electric force in bringing a test charge from \infty to r is given by 

                    W=\int_{\infty}^{r}\frac{KQq}{x^2}dx=-\frac{KQq}{r}

    And negative of this work done is equal to electric potential energy

               So U=\frac{KQq}{r}

U \rightarrow electric potential energy 

 r \rightarrow distance between two

Change of potential energy-

if a charge q  is moved from  r_{1 } to r_{2 } in a electric field produced by charge Q

Then Change of potential energy is given as

 \Delta U=KQq\left [ \frac{1}{r_{2}}-\frac{1}{r_{1}} \right ]

 

\Delta U \rightarrow change of energy

r_{1},r_{2}\rightarrow distances

Potential Energy Of System Of two Charge-

 

U=\frac{KQ_{1}Q_{2}}{r}  \left ( S.I \right ) where K=\frac{1}{4\pi \epsilon _{0}}

   Potential Energy For a system of 3 charges-

9U= K\left ( \frac{Q_{1}Q_{2}}{r_{12}} +\frac{Q_{2}Q_{3}}{r_{23}}+\frac{Q_{1}Q_{3}}{r_{13}}\right )

Work energy relation-

W= U_{f}-U_{i}

Where W=work done by an external force

U_{f}- \, final\, P.E

U_{i}-\, initial\, P.E.

The relation between Potential and Potential energy-

As U=\frac{KQq}{r}=q\left [ \frac{KQ}{r} \right ]

But V=\frac{KQ}{r}  

So U=qV 

Or potential is defined as Potential energy Per unit charge.

i.e V =\frac{W}{Q}=\frac{U}{Q}

Where V\rightarrow Potential

U\rightarrow Potential energy

Electron Volt-

1\, e v= 1.6\times 10^{-19}J=1.6\times 10^{-12}erg.

It is the smallest practical unit of energy that is used in atomic and nuclear physics.

Electric potential Energy of Uniformly charged sphere (Self-energy of the Uniformly charged sphere)

U=\frac{3Q^{2}}{20\pi \epsilon _{0}R}

Electric potential Energy of Uniformly charged hollow sphere (Self-energy of the Uniformly charged hollow sphere)

\dpi{100} U=\frac{Q^{2}}{8\pi \epsilon _{0}R}

Where R - Radius and Q - total charge.

Energy density- It is defined as the energy stored for unit volume.

U_{v}=\frac{U}{V}

Where U-Potential\, Energy and  V-Volume.

 

 

 

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Electrostatic Potential energy

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