VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Electrostatic Potential energy is considered one of the most asked concept.
62 Questions around this concept.
In free space, a particle A of charge $1 \mu C$ is held fixed at a point P. Another particle B of the same charge and mass of $4 \mu \mathrm{~kg}$ is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is:
$$
\left[\text { Take } \frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right]
$$
Two equal charges are placed at a distance of
and a third charge
is placed at the midpoint. The potential energy of the system is :
If a person brings an electron towards another electron then the change in potential energy of the entire system will be
JEE Main 2025: City Slip Link | Study Plan | Official Question Paper (Session 1)
JEE Main 2025: Sample Papers | Mock Tests | PYQs | High Scoring Topics | College Predictor
New: Meet Careers360 experts in your city and get guidance on shortlisting colleges
Apply to TOP B.Tech /BE Entrance exams: VITEEE | MET | AEEE | BITSAT
When a charged particle is placed in uniform electric field its momentum is p. If the mass of particle is doubled, then the momentum of particle will become:
Consider an initially neutral hollow conducting spherical shell with an inner radius r and an outer radius of 2r. A point charge +Q is now placed inside the shell at a distance r/2 from the center. The shell is then grounded by connecting the outer surface to the earth. P is an external point at a distance of 2r from the point charge +Q on the line passing through the center and the point charge +Q as shown in the figure.
The magnitude of the force on a test charge +q placed at P will be
One can define alpha-Volt $(aV)$ to be the energy acquired by a particle when it is accelerated by a potential of 1 Volt. For this problem, you may take a proton to be 2000 times heavier than an electron. Then:
A molecule with a dipole moment p is placed in an electric field of strength E. Initially the dipole is aligned parallel to the field. If the dipole is to be rotated to be anti-parallel to the field, the work required to be done by an external agency is
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements | Last Date to Apply: 28th March
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: $\left[M L^2 T^{-2}\right]$
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 $F=\frac{K Q q}{r^2}$
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{K Q q}{x^2} d x=-\frac{K Q q}{r}
$
And negative of this work done is equal to electric potential energy
$
\text { So } U=\frac{K Q q}{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=K Q q\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{K Q_1 Q_2}{r}(S . I)_{\text {where }} K=\frac{1}{4 \pi \epsilon_0}
$
Potential Energy For a system of 3 charges-
$
U=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
$
\begin{aligned}
& U_f-\text { final P.E } \\
& U_i-\text { initial P.E. }
\end{aligned}
$
The relation between Potential and Potential energy-
$
\begin{aligned}
& U=\frac{K Q q}{r}=q\left[\frac{K Q}{r}\right] \\
& \text { As } \\
& \text { But }=\frac{K Q}{r}
\end{aligned}
$
So $U=q V$
Or potential is defined as Potential energy Per unit charge.
$
\text { i.e } V=\frac{W}{Q}=\frac{U}{Q}
$
Where $V \rightarrow$ Potential
$U \rightarrow$ Potential energy
Electron Volt-
$
1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}=1.6 \times 10^{-12} \mathrm{erg}
$
It is the smallest practical unit of energy which is used in atomic and nuclear physics.
Electric potential Energy of Uniformly charged sphere-
$
U=\frac{3 Q^2}{20 \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.
"Stay in the loop. Receive exam news, study resources, and expert advice!"