MHT-CET Physics · Electrostatics

Electric Potential and Potential Energy

Potential is the work per unit charge to bring a charge in from infinity; it is a scalar, so potentials from several charges simply add, and the energy of a group of charges is the sum over every pair.

All Electrostatics notes MHT-CET Physics strategy

Why this matters

43 PYQs, the largest page in the chapter, nine of them HARD and spread across all five shapes below. Five shapes repeat: the potential of charges at the corners of a shape, the field from a potential function, the energy of a group, the speed a charge gains through a potential difference, and drops or spheres that merge or share charge.

Concept 1 of 5: Potential of Point Charges, and Equipotentials

Potential has no direction, so the potential of several charges is just the sum of kq/r, with the signs of the charges kept. Points at the same distance from a charge share a potential, and moving a charge between points of equal potential costs no work at all.

Definition

$V = \dfrac{kq}{r}$ ; for many charges $V = k\sum \dfrac{q_i}{r_i}$ . Negative charges give negative terms.
Regular polygon with $n$ equal charges $q$ and centre-to-vertex distance $r$ : $V = \dfrac{nkq}{r}$ . A regular hexagon's centre-to-vertex distance EQUALS its side.
Zero potential between unlike charges $q_1$ and $-q_2$ a distance $d$ apart: $\dfrac{q_1}{x} = \dfrac{q_2}{d - x}$ .
Equipotential surface: $W = q\Delta V = 0$ for any move on it — along an arc round a point charge, or between the corners of a square with a charge at its centre.
Charged conducting sphere: $E = 0$ inside, $V = \dfrac{kQ}{R}$ everywhere inside. Half ring of density $\lambda$ : $V = \dfrac{\lambda}{4\varepsilon_0}$ at the centre, whatever the radius.

Potential of charges

V = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}

Worked example

Charges

4\,\mu

C and

-6\,\mu

C are 50 cm apart. Where between them is the potential zero?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q4Moderate

Example 1 · Electrostatics · Electric Potential and Potential Energy

A regular hexagon of side 10 cm has a charge

1\,\mu\text{C}

at each of its vertices. The potential at the centre of hexagon is

\left[\frac{1}{4\pi\varepsilon_{0}}=9\times10^{9}\text{ SI unit}\right]

Dropping the sign of a negative charge

Potentials add with their signs. Four charges

+q, +q, -q, -q

on a square give

\frac{2kq}{L}\left(1 - \frac{1}{\sqrt{5}}\right)

at the midpoint of the positive side; the

1 + \frac{1}{\sqrt{5}}

option is what you get by adding magnitudes.

Drill 13 more on potential of point charges, and equipotentials

Concept 2 of 5: Field From Potential, and Potential From Field

The field points the way the potential falls fastest, and its size is how steeply it falls. So the field is minus the slope of V, and the potential difference is minus the area under E. In a uniform field that is just V = Ed.

Definition

$E = -\dfrac{dV}{dx}$ . $V = 4x^2 + 8x - 3$ gives $E = -(8x + 8)$ .
$V_B - V_A = -\displaystyle\int_A^B E\,dx$ . For $E = 30x^2$ : $V(2) - V(0) = -80$ V.
Uniform field: $V = Ed$ , and $E = \dfrac{F}{q}$ .
Work by an outside agent to move $q$ from $A$ to $B$ : $W = q(V_B - V_A)$ , so $V_B = \dfrac{W}{q} + V_A$ .
Inside a sphere with $V = ar^2 + b$ : $E = -2ar$ , and Gauss's law gives $\rho = -6a\varepsilon_0$ .

Field and potential

E = -\frac{dV}{dx}, \qquad V_B - V_A = -\int_A^B E\,dx

Worked example

V = 3x^2 - 2x + 5

volt (x in metres). Field at

x = 1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q10Easy

Example 2 · Electrostatics · Electric Potential and Potential Energy

The electric potential '

V

' is given as a function of distance '

x

' (metre) by

V=\left( 4x^{2}+ 8x- 3 \right)V

. The value of electric field at

x= 0.5\text{ }m

, in

V/m

Forgetting the minus sign

E = -\frac{dV}{dx}

. Where

V

rises with

x

the field points along

-x

. The paper always offers the same magnitude with both signs or both directions.

Drill 6 more on field from potential, and potential from field

Concept 3 of 5: Potential Energy of a Group of Charges

The energy of a group is the work to assemble it, and that work is the sum of kq_iq_j/r over every PAIR, counted once. Setting the total to zero is the commonest HARD question here: write one term per pair and solve for the unknown charge.

Definition

Two charges: $U = \dfrac{kq_1q_2}{r}$ . Three charges: three pair terms; four charges: six.
Equilateral triangle of side $a$ with $q_1, q_2, Q$ : $U = 0 \Rightarrow q_1q_2 + Q(q_1 + q_2) = 0$ .
Right isosceles triangle: legs $a$ , hypotenuse $\sqrt{2}a$ — read the figure for which pair sits on the hypotenuse.
Moving $q_2$ a distance $x$ closer to $q_1$ (from $d$ ): $\Delta U = kq_1q_2\left(\dfrac{1}{d - x} - \dfrac{1}{d}\right) = \dfrac{kq_1q_2x}{d(d - x)}$ .
A single charge in an external potential: $U = qV$ ; a positive charge moved to higher potential gains energy.

Energy of a system

U = \sum_{\text{pairs } i<j} \frac{kq_iq_j}{r_{ij}}

Worked example

Charges of 1, 2 and

3\,\mu

C sit at the corners of an equilateral triangle of side 10 cm. Energy of the system?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q44Moderate

Example 3 · Electrostatics · Electric Potential and Potential Energy

Three point charges

+q, +2q

and

+Q

are placed at the three vertices of an equilateral triangle. If the potential energy of the system of three charges is zero, the value of Q in terms of q is

Missing a pair

Three charges have THREE pairs, including the two outer charges with each other.

q, -2q, q

on a line of length

2r

gives

-\frac{7q^2}{8\pi\varepsilon_0 r}

only with the

\frac{q^2}{2r}

end-to-end term.

Drill 9 more on potential energy of a group of charges

Concept 4 of 5: Speed Gained Through a Potential Difference

A charge that falls through a potential difference V turns qV of potential energy into kinetic energy. The speed therefore goes as the square root of q/m — twice the charge on the same mass gives √2 times the speed, not twice.

Definition

$qV = \dfrac{1}{2}mv^2 \Rightarrow v = \sqrt{\dfrac{2qV}{m}}$ , so $v \propto \sqrt{\dfrac{q}{m}}$ .
In a uniform field over a distance $l$ : $qEl = \dfrac{1}{2}mv^2$ , so $v = \sqrt{\dfrac{2qEl}{m}}$ and $\dfrac{q}{m} = \dfrac{v^2}{2El}$ .
Energy in electron-volts: a charge $ne$ through $V$ volts gains $nV$ eV.
Two equal charges released together share the change in their mutual energy equally: $2 \cdot \dfrac{1}{2}mv^2 = |\Delta U|$ .

Energy gained

qV = \tfrac{1}{2}mv^2, \qquad v = \sqrt{\frac{2qV}{m}}

Worked example

An alpha particle (charge

2e

, mass

4m

) and a proton (

e

m

) fall through the same potential difference. Ratio of their speeds?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q32Easy

Example 4 · Electrostatics · Electric Potential and Potential Energy

Two particles A and B having same mass have charge

+q

and

+4q

, respectively. When they are allowed to fall from rest through the same electric potential difference, the ratio of their speeds

v_A

v_B

will become

Taking the charge ratio as the speed ratio

+q

and

+4q

of the same mass through the same V: the ENERGIES are 1 : 4, the speeds 1 : 2. The 1 : 4 option is always there.

Drill 4 more on speed gained through a potential difference

Concept 5 of 5: Merging Drops and Connected Spheres

When n charged drops merge, the charge goes up n times but the radius only n^{1/3} times, so the potential kQ/R rises as n^{2/3}. When two spheres are joined by a wire, charge flows until their potentials match — so each ends up with charge in proportion to its radius, and the smaller one has the denser surface charge.

Definition

$n$ identical drops, each at potential $v$ : $R = n^{1/3}r$ , $Q = nq$ , $V = n^{2/3}v$ . 27 drops give 9 times the potential; 1000 drops, 100 times.
Spheres in contact or joined: $\dfrac{q_1}{r_1} = \dfrac{q_2}{r_2}$ , so charge $\propto$ radius and $\sigma \propto \dfrac{1}{r}$ .
Two spheres at the same potential: $\dfrac{\sigma_1}{\sigma_2} = \dfrac{C_1 R_2^2}{C_2 R_1^2}$ .
Van de Graaff generator rests on corona discharge (spraying charge from sharp points), on charge given to a hollow conductor moving to its outer surface, and on the potential of an isolated conductor rising as charge is added. It does NOT use crossed electric and magnetic fields.

Merging drops

V_{\text{big}} = n^{2/3}\,v

Worked example

Eight identical drops each charged to 10 V merge into one. Potential of the big drop?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q23Moderate

Example 5 · Electrostatics · Electric Potential and Potential Energy

Assuming the drops to be spherical, 27 identical drops of mercury are charged simultaneously to the same potential of 20 volt. If all the charged drops are made to combine to form one big drop, then potential of big drop will be

Potential rising as n, or as the cube root of n

Charge rises

n

times and radius

n^{1/3}

times; the potential is their ratio,

n^{2/3}

. Both

n

and

n^{1/3}

are offered every time.

Drill 6 more on merging drops and connected spheres

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Potential of Point Charges, and Equipotentials
Potential of charges
$V = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}$
Field From Potential, and Potential From Field
Field and potential
$E = -\frac{dV}{dx}, \qquad V_B - V_A = -\int_A^B E\,dx$
Potential Energy of a Group of Charges
Energy of a system
$U = \sum_{\text{pairs } i<j} \frac{kq_iq_j}{r_{ij}}$
Speed Gained Through a Potential Difference
Energy gained
$qV = \tfrac{1}{2}mv^2, \qquad v = \sqrt{\frac{2qV}{m}}$
Merging Drops and Connected Spheres
Merging drops
$V_{\text{big}} = n^{2/3}\,v$

Watch out for (5)

Dropping the sign of a negative charge
→ Potential of Point Charges, and Equipotentials
Forgetting the minus sign
→ Field From Potential, and Potential From Field
Missing a pair
→ Potential Energy of a Group of Charges
Taking the charge ratio as the speed ratio
→ Speed Gained Through a Potential Difference
Potential rising as n, or as the cube root of n
→ Merging Drops and Connected Spheres

Test yourself on Electrostatics

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