krit.club logo

Electrostatics - Electrostatic Potential

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Electric Potential (VV): The amount of work done in bringing a unit positive test charge from infinity to a point in an electric field. It is a scalar quantity, and its SI unit is the Volt (VV), where 1 V=1 J/C1 \text{ V} = 1 \text{ J/C}.

•

Potential Difference (ΔV\Delta V): The work done per unit charge in moving a positive test charge from one point (say AA) to another (say BB). Given by VB−VA=WABq0V_B - V_A = \frac{W_{AB}}{q_0}.

•

Conservative Nature: The work done in moving a charge in an electrostatic field depends only on the initial and final positions and is independent of the path taken.

•

Equipotential Surface: A surface at every point of which the electric potential is the same. No work is done in moving a charge along an equipotential surface (dW=0dW = 0). Electric field lines are always perpendicular to equipotential surfaces.

•

Relation between Electric Field and Potential: The electric field (EE) is the negative gradient of electric potential, expressed as E=−dVdrE = -\frac{dV}{dr}. The negative sign indicates that the direction of the electric field is in the direction of decreasing potential.

•

Electrostatic Potential Energy (UU): The energy stored in a system of charges due to their configuration. For a system of two charges, it is the work done to bring them from infinity to their current positions.

•

Potential due to an Electric Dipole: Unlike a point charge where V∝1rV \propto \frac{1}{r}, for a dipole, the potential at large distances varies as V∝1r2V \propto \frac{1}{r^2}. At any point on the equatorial line, the potential is always zero.

📐Formulae

V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}

VB−VA=−∫ABE⃗⋅dr⃗V_B - V_A = -\int_{A}^{B} \vec{E} \cdot d\vec{r}

V=14πϵ0pcos⁡θr2V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2}

E=−dVdrE = -\frac{dV}{dr}

U=14πϵ0q1q2r12U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r_{12}}

W=q(VB−VA)W = q(V_B - V_A) (Work done in moving a charge qq)

U=−p⃗⋅E⃗=−pEcos⁡θU = -\vec{p} \cdot \vec{E} = -pE \cos \theta (Potential energy of a dipole in an external field)

💡Examples

Problem 1:

Calculate the electric potential at a point 10 cm10 \text{ cm} away from a point charge of 2×10−7 C2 \times 10^{-7} \text{ C}.

Solution:

Given: q=2×10−7 Cq = 2 \times 10^{-7} \text{ C}, r=10 cm=0.1 mr = 10 \text{ cm} = 0.1 \text{ m}. Using the formula V=14πϵ0qrV = \frac{1}{4\pi\epsilon_0} \frac{q}{r}, where 14πϵ0=9×109 Nm2/C2\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2: V=9×109×2×10−70.1=1.8×104 VV = 9 \times 10^9 \times \frac{2 \times 10^{-7}}{0.1} = 1.8 \times 10^4 \text{ V}

Explanation:

The potential is calculated by applying the point charge formula directly, ensuring the distance is converted to SI units (meters).

Problem 2:

Determine the work done in moving a charge of 5 \muC5 \text{ \mu C} between two points having a potential difference of 12 V12 \text{ V}.

Solution:

Given: q=5×10−6 Cq = 5 \times 10^{-6} \text{ C}, ΔV=12 V\Delta V = 12 \text{ V}. Work done W=qΔVW = q \Delta V. W=(5×10−6 C)×(12 V)=60×10−6 J=6×10−5 JW = (5 \times 10^{-6} \text{ C}) \times (12 \text{ V}) = 60 \times 10^{-6} \text{ J} = 6 \times 10^{-5} \text{ J}

Explanation:

Work done is the product of the magnitude of the charge and the potential difference through which it is moved.

Problem 3:

Two charges 3×10−8 C3 \times 10^{-8} \text{ C} and −2×10−8 C-2 \times 10^{-8} \text{ C} are located 15 cm15 \text{ cm} apart. At what point on the line joining the two charges is the electric potential zero?

Solution:

Let the point PP be at a distance xx from the positive charge. For V=0V = 0: 14πϵ0[3×10−8x+−2×10−815−x]=0\frac{1}{4\pi\epsilon_0} \left[ \frac{3 \times 10^{-8}}{x} + \frac{-2 \times 10^{-8}}{15 - x} \right] = 0 3x=215−x⇒45−3x=2x⇒5x=45⇒x=9 cm\frac{3}{x} = \frac{2}{15 - x} \Rightarrow 45 - 3x = 2x \Rightarrow 5x = 45 \Rightarrow x = 9 \text{ cm}

Explanation:

The net potential is the algebraic sum of potentials due to individual charges. We set this sum to zero and solve for the distance.

Electrostatic Potential - Revision Notes & Key Formulas | ICSE Class 12 Physics