Electrostatics Concept Page - 11

Example

Energy required in changing geometrical configuration of a charged body

Example:
An isolated metal sphere of radius r$r$ is given a charge q$q$. Calculate the energy required to change the radius to r/2$r/2$ and same charge.Solution:
First, finding the potential energy in creating the sphere,
Let at any instant q$q$ be the charge on the sphere
Now potential of the sphere =kqr$={\displaystyle \frac{kq}{r}}$
Work done to add a dq$dq$ charge to this sphere
dw=kqr⋅dq$dw={\displaystyle \frac{kq}{r}}\cdot dq$     (As V∞=0$V_{\mathrm{\infty }}=0$)
∴$\therefore$ total work done to accumulate q$q$ charge is
∫dw=∫q0kqr⋅dq$\int dw={\int }_0^q{\displaystyle \frac{kq}{r}}\cdot dq$
⇒w=kr[q22]q0$\Rightarrow w={\displaystyle \frac{k}{r}}[{\displaystyle \frac{q^2}{2}}{]}_0^q$
⇒w=kq22r$\Rightarrow w={\displaystyle \frac{k{q}^2}{2r}}$
⇒w=q24πε0r⋅2$\Rightarrow w={\displaystyle \frac{q^2}{4\pi {\epsilon }_{0}r\cdot 2}}$
⇒w=q28πε0r$\Rightarrow w={\displaystyle \frac{q^2}{8\pi {\epsilon }_{0}r}}$
Potential energy of the sphere initially is Ui=q28πε0r$U_i={\displaystyle \frac{q^2}{8\pi {\epsilon }_{0}r}}$
Potential energy of the sphere finally is Uf=q24πε0r$U_f={\displaystyle \frac{q^2}{4\pi {\epsilon }_{0}r}}$
Energy required to change the configuration, W=Uf−Ui=q28πε0r$W=U_f-U_i={\displaystyle \frac{q^2}{8\pi {\epsilon }_{0}r}}$

Formula

Potential due to a charged non-conducting sphere

V=14πϵ0Q(3R2−r2)2R3$V={\displaystyle \frac{1}{4\pi {ϵ}_0}}{\displaystyle \frac{Q(3R^2-r^2)}{2R^3}}$             (r<R)
V=kQR$V={\displaystyle \frac{kQ}{R}}$            (r=R)
V=kQr$V={\displaystyle \frac{kQ}{r}}$              (r>R)
where k=14πϵ0$k={\displaystyle \frac{1}{4\pi {ϵ}_0}}$, R$R$ is the radius of the sphere and r$r$ is the distance from the centre

Diagram

Plot of potential for a charged non-conducting sphere

Example

Potential of a uniformly charged cylinder

Example:
Calculate the electric potential due to an infinitely long uniformly charged cylinder with charge density ρo${\rho }_o$ and radius R$R$, inside and outside the cylinder. Assume potential at axis is zero.
Solution:
For r<R$r<R$,
      Electric field using Gauss Law, E=ρr2εo$E={\displaystyle \frac{\rho r}{2{\epsilon }_o}}$
      Electric potential, dV=−E⃗ .d⃗ r$dV=-\overrightarrow{E}.\overrightarrow{d}r$
                                  Vr−V0=−∫r0ρr2εodr${\displaystyle V_r-V_0=-{\int }_0^r{\displaystyle \frac{\rho r}{2{\epsilon }_o}}dr}$
                                  Vr−0=−ρr24εo$V_r-0=-{\displaystyle \frac{\rho r^2}{4{\epsilon }_o}}$
For r=R,VR=−ρR24εo$r=R,V_R=-{\displaystyle \frac{\rho R^2}{4{\epsilon }_o}}$
For r>R$r>R$,
      Electric field using Gauss Law, E=ρR22rεo$E={\displaystyle \frac{\rho R^2}{2r{\epsilon }_o}}$
      Electric potential, dV=−E⃗ .d⃗ r$dV=-\overrightarrow{E}.\overrightarrow{d}r$
                                  Vr−VR=−∫rRρR22rεodr${\displaystyle V_r-V_R=-{\int }_R^r{\displaystyle \frac{\rho R^2}{2r{\epsilon }_o}}dr}$
                                  Vr=−ρR24εo−ρR22εoln(rR)$V_r=-{\displaystyle \frac{\rho R^2}{4{\epsilon }_o}}-{\displaystyle \frac{\rho R^2}{2{\epsilon }_o}}ln\left({\displaystyle \frac{r}{R}}\right)$

Formula

Discuss and explain the relation between electric field intensity and potential

The relationship between electric field E$E$ and scalar potential ϕ$\varphi$ is given as:
E=▽ϕ$E=▽\varphi$, where ▽=$▽=$gradient operator.

Example

Integral relationship between electric field and potential

The potential at the origin is zero due to electric field E¯=20i^+30j^NC−1$\overline{E}=20\hat{i}+30\hat{j}N{C}^{-1}$ . The potential at point P(2m, 2m) is:

We know Vr1−Vr2=−∫r1r2Edr$V_{r_1}-V_{r_2}=-{\int }_{r_2}^{r_1}Edr$
Vx0−Vx2=−∫0220dr$V_{x_0}-V_{x_2}=-{\int }_2^{0}20dr$
⇒Vx2=−40$\Rightarrow V_{x_2}=-40$
Vy0−Vy2=−∫0230dr$V_{y_0}-V_{y_2}=-{\int }_2^{0}30dr$
Vy2=−60$V_{y_2}=-60$
total V=Vx2+Vy2=−100V$totalV=V_{x_2}+V_{y_2}=-100V$

Example

Differential relationship between electric field and potential

In the figure shown, the electric field intensity at r=1 m$r=1m$, r=6 m$r=6m$, r=9 m$r=9m$ in Vm−1$V{m}^{-1}$ is:We know: E=−dVdr$E={\displaystyle \frac{-dV}{dr}}$
Given plot shows the variation of V$V$ against r$r$.
∴dVdr$\therefore {\displaystyle \frac{dV}{dr}}$ is the slope of graph
Since, slope at (1,5)$(1,5)$ is 5$5$, therefore E|r=1 m=−5$E{|}_{r=1m}=-5$ Since, slope at (6,10)$(6,10)$ is 0$0$, therefore E|r=6 m=0$E{|}_{r=6m}=0$Since, slope at (9,5)$(9,5)$ is −5$-5$, therefore  Er=9 m=5$E_{r=9m}=5$

Example

Electric field due to a system of point charges

An infinite number of charges each of magnitude q$q$ are placed on x - axis at distances of 1,2, 4, 8, ... meter from the origin. The intensity of the electric field at origin is:We know E=kqr2$E={\displaystyle \frac{kq}{r^2}}$
Now electric field at origin is given as Eo=kq12+kq22+kq42+kq82+−−−−$E_o={\displaystyle \frac{kq}{1^2}}+{\displaystyle \frac{kq}{2^2}}+{\displaystyle \frac{kq}{4^2}}+{\displaystyle \frac{kq}{8^2}}+----$
=kq(1+122+124+126+−−−−)$=kq(1+{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{2^6}}+----)$
We know in G. P sum to infinite =a1−r$={\displaystyle \frac{a}{1-r}}$
=kq⋅(11−14)$=kq\cdot ({\displaystyle \frac{1}{1-{\displaystyle \frac{1}{4}}}})$
=q4πε0⋅43=q3πε0.$={\displaystyle \frac{q}{4\pi {\epsilon }_0}}\cdot {\displaystyle \frac{4}{3}}={\displaystyle \frac{q}{3\pi {\epsilon }_0.}}$

Diagram

Gravitational Equipotential Surface

Gravitational equipotential surface is the locus of points which are at the same potential. Electric field in a space is always perpendicular to the equipotential surfaces and zero along the plane of equipotential surfaces.
Example:
For a point charge, V=Q4πεoR$V={\displaystyle \frac{Q}{4\pi {\epsilon }_{o}R}}$
V=$V=$ constant, ⟹R=$⟹R=$ constant.
Hence for a point charge, equipotential surfaces are concentric spheres centered at the point charge.

Definition

Equipotential surfaces

Surfaces having same potential are termed as equipotential surfaces
The properties of equipotential surfaces can be summarized as follows:

• The electric field lines are normal to the equipotentials and are directed from higher to lower potentials. 
• By symmetry, the equipotential surfaces produced by a point charge form a family of concentric spheres, and for a constant electric field, a family of planes normal to the field lines. 
• The tangential component of the electric field along the equipotential surface is zero, otherwise non-vanishing work would be done to move a charge from one point on the surface to the other.
• Work done in moving a particle along an equipotential surface is zero. 

BookMarks

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