Equipotential Surfaces

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   If the field were not normal to the equipotential surface, it would have a non-zero component along the surface, and moving a test charge against this component would require work. But no work is required to move a test charge between any two points on an equipotential surface (potential difference is zero there). Hence the electric field must be normal to the equipotential surface at every point.

2. 2.↑ back to the question

   (a) Since the field is uniform along $z$, the equipotential surfaces are planes perpendicular to the $z$-axis (parallel $xy$-planes), equally spaced. (b) Let $P$ be an axial point at distance $r$ from the centre of a dipole of charges $-q$ (distance $r+a$) and $+q$ (distance $r-a$): $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{(-q)}{r+a}+\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r-a}=\dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{r-a}-\dfrac{1}{r+a}\right]=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2-a^2}$ (using $p=q(2a)$). For $r\gg a$: $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2}$.

3. 3.↑ back to the question

   The equipotential surfaces due to an electric dipole are closed, non-concentric, non-uniformly spaced surfaces, symmetric about the dipole axis and denser near the charges: field lines run from $+q$ to $-q$, and the equipotential surfaces are perpendicular to the field lines at every point (positive-potential surfaces cluster around $+q$, negative-potential surfaces around $-q$, with a plane of zero potential midway, perpendicular to the axis).

4. 4.↑ back to the question

   An equipotential surface is a surface on which the electric potential has the same value at every point. (i) For a single point charge: concentric spheres centred on the charge. (ii) For a constant field along $Z$: parallel planes perpendicular to the $Z$-axis. Equipotential surfaces about a single charge are not equidistant because $V\propto1/r$, so the surfaces get closer together as $r$ decreases (spacing is not linear in $r$). (iii) No. If the field had a tangential component, work would be done in moving a charge along the surface, but $W_{AB}=q_0(V_B-V_A)=0$ on an equipotential surface. Hence the field cannot exist tangential to an equipotential surface.

5. 5.↑ back to the question

   If $\vec E$ is the field at $P$ due to charge $+q$ at $O$, a test charge $q_0$ experiences force $q_0\vec E$; the external force needed to move it against the field is $\vec F=-q_0\vec E$. Work done in a small displacement $\overline{PQ}=\delta l$ (against the field, i.e. towards $A$): $\Delta W=\vec F\cdot\vec{\delta l}=(-q_0\vec E)\cdot\vec{\delta l}=-q_0E\,\delta l\cos180°=q_0E\,\delta l$. As distance $r$ from $O$ decreases in the direction of $\delta l$: $\delta W=-q_0E\,\delta r \Rightarrow \dfrac{\delta W}{q_0}=-E\,\delta r$ ...(ii) Also $\delta V=\dfrac{\delta W}{q_0}$ ...(i). From (i) and (ii): $\delta V=-E\,\delta r$, i.e. $E=-\dfrac{\delta V}{\delta r}$. Hence electric field at a point is equal to the negative gradient of the electrostatic potential at that point. Conclusions: (i) No work is done in moving a test charge over an equipotential surface. (ii) The electric field is always at right angles to the equipotential surface.

6. 6.↑ back to the question

   The equipotential surfaces are concentric spheres centred on $-q$, spaced more closely near the charge (since $V\propto1/r$) and further apart at large distances (weaker field). The electric field lines point radially inward, towards $-q$, perpendicular to every equipotential surface at each point.

7. 7.↑ back to the question

   (i) $\vec E=-\dfrac{\partial V}{\partial x}\hat i=-\dfrac{d}{dx}(10x+5)\hat i=-10\hat i\,\text{V/m}$; magnitude $10\,\text{V/m}$, directed along $-x$. (ii) $\phi=\displaystyle\oint\vec E\cdot d\vec S$. $\vec E$ (uniform, along $-x$) contributes flux only through the two faces perpendicular to $x$, each of area $(0.2)^2\,\text{m}^2$: $\phi=0+0+(+10)(0.2)^2+(-10)(0.2)^2+0+0=0$ (flux entering one face exactly cancels flux leaving the opposite face, since $E$ is uniform).

8. 8.↑ back to the question

   (i) No work is done in moving a charge from one point to another on an equipotential surface. (ii) The electric field is always perpendicular to the equipotential surface at every point (and two equipotential surfaces can never intersect).

9. 9.↑ back to the question

   $V=-\displaystyle\int E\,dr=-\int_1^{10}(10r+5)\,dr=-\left[5r^2+5r\right]_1^{10}=-[(500+50)-(5+5)]=-540\,\text{V}$. So the potential decreases by 540 V on moving from $r=1\,\text{m}$ to $r=10\,\text{m}$.

10. 10.↑ back to the question

    Using $E=-\dfrac{dV}{dx}$: for $x=0$ to $1$ ($V=kx$, rising): $E=-k$ (constant, negative). For $x=1$ to $2$ ($V=k$, constant): $E=0$. For $x=2$ to $3$ ($V=-kx+3k$, falling): $E=+k$ (constant, positive). The $E$-$x$ graph is a rectangular step: $-k$ from $0$ to $1$, zero from $1$ to $2$, and $+k$ from $2$ to $3$.

11. 11.↑ back to the question

    No. For example, inside a uniformly charged spherical shell, the electric field is zero everywhere, but the potential is constant and non-zero (equal to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}$, same as on the surface) throughout the interior.

12. 12.Answer: (A)↑ back to the question

    $E=-\dfrac{dV}{dx}=-\dfrac{d}{dx}(3x^2)=-6x$. At $x=1$: $E=-6\,\text{V m}^{-1}$, i.e. magnitude $6\,\text{V m}^{-1}$ along $-x$-axis.

13. 13.Answer: (A)↑ back to the question

    Correct option: (a). At a large distance the charge collection behaves like a single point charge of magnitude equal to the net charge, so the equipotentials are concentric spheres (since $V\propto1/r$).