Potential due to a Point Charge

Potential due to a Point Charge

In the previous section, we defined electrostatic potential as the work done per unit charge in bringing a charge from infinity to a point. Now, we will mathematically derive the exact expression for the potential at a distance $r$ from a single isolated point charge $Q$. This is the fundamental 'building block' formula for electrostatics.

Derivation of Potential ($V$)

1. The Setup:

Consider a point charge $+Q$ placed at the origin $O$. We want to find the electrostatic potential at a point $P$ at a distance $r$ from the origin. To do this, we calculate the work done in bringing a unit positive test charge ($q_0 = +1$ C) from infinity to point $P$.

2. Force on Unit Charge:

At some intermediate point $P'$ on the path, at a distance $r'$ from the origin, the electrostatic force on the unit positive charge is given by Coulomb's Law: $$F = \frac{1}{4\pi\epsilon_0} \frac{Q \cdot 1}{(r')^2}$$ The direction of this force is away from the origin (repulsive).

3. Small Work Done ($dW$):

Let the unit charge be moved a very small distance $dr'$ towards the origin (from $P'$ to a slightly closer point). The small work done $dW$ against the electrostatic force is: $$dW = \vec{F} \cdot d\vec{r}' = F dr' \cos(180^{\circ})$$ Since the displacement $d\vec{r}'$ is opposite to the direction of force $\vec{F}$: $$dW = -F dr' = -\frac{Q}{4\pi\epsilon_0 (r')^2} dr'$$

4. Total Work Done ($W$):

To find the total work done in moving the unit charge from infinity ($r' = \infty$) to the point $P$ ($r' = r$), we integrate the small work: $$W = \int_{\infty}^{r} -\frac{Q}{4\pi\epsilon_0 (r')^2} dr'$$ $$W = -\frac{Q}{4\pi\epsilon_0} \int_{\infty}^{r} (r')^{-2} dr'$$

5. Solving the Integration:

Using the power rule for integration $\int x^n dx = \frac{x^{n+1}}{n+1}$: $$W = -\frac{Q}{4\pi\epsilon_0} \left[ \frac{(r')^{-1}}{-1} \right]_{\infty}^{r}$$ $$W = \frac{Q}{4\pi\epsilon_0} \left[ \frac{1}{r'} \right]_{\infty}^{r}$$ $$W = \frac{Q}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{\infty} \right)$$ Since $1/\infty = 0$: $$W = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$$

6. Final Formula:

By definition, the work done in bringing a unit positive charge from infinity to point $P$ is the electrostatic potential $V$ at that point. $$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$$

Graphical Variation of $V$ and $E$ with $r$

It is crucial to compare how Potential ($V$) and Electric Field ($E$) change with distance:

• Electric Field: $E \propto 1/r^2$ (Follows inverse square law).
• Potential: $V \propto 1/r$ (Follows inverse law).

Key Observations

• Distance Relationship: Unlike the electric field which follows $1/r^2$, the potential follows a $1/r$ relationship. This means potential decreases more slowly as we move away from the charge.
• Sign Dependency: The potential is a scalar. For a positive charge, $V$ is positive; for a negative charge, $V$ is negative.
• Graph comparison: A plot of $V$ vs $r$ and $E$ vs $r$ clearly shows that the $E$ curve is steeper than the $V$ curve.

🧠 Memory Capsule

• The Formula: $V = kQ/r$. Potential is inversely proportional to distance.
• Scalar Nature: Potential is a scalar. When calculating total potential from multiple charges, simply use algebraic addition (no vectors needed!).
• Sign Matters: If $Q$ is positive, $V$ is positive. If $Q$ is negative, $V$ is negative. Always substitute the sign of the charge in the formula.
• Infinity Reference: By convention, potential at infinity is taken to be zero ($V_{\infty} = 0$).

Example 1: Basic Potential Calculation

Calculate the electric potential at a point 9 cm away from a point charge of $4 \times 10^{-7}$ C.

Solution:

1. Given: $Q = 4 \times 10^{-7}$ C, $r = 9$ cm $= 0.09$ m.
2. Formula: $$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = k\frac{Q}{r}$$
3. Substitution: $$V = 9 \times 10^9 \times \frac{4 \times 10^{-7}}{0.09}$$
4. Calculation: $$V = \frac{36 \times 10^2}{0.09} = 4 \times 10^4 \text{ V}$$
5. Final Answer: $$V = 4 \times 10^4 \text{ V}$$

Example 2: Potential at a Point (Negative Charge)

What is the potential at a distance of 0.5 m from a point charge of $-2 \mu C$?

Solution:

1. Given: $Q = -2 \times 10^{-6}$ C, $r = 0.5$ m.
2. Formula: $$V = k\frac{Q}{r}$$
3. Substitution: $$V = 9 \times 10^9 \times \frac{-2 \times 10^{-6}}{0.5}$$
4. Calculation: $$V = -36 \times 10^3 \text{ V}$$
5. Final Answer: $$V = -3.6 \times 10^4 \text{ V}$$

Example 3: Distance for a Given Potential

At what distance from a $1 \mu C$ charge is the potential 900 V?

Solution:

1. Given: $Q = 1 \times 10^{-6}$ C, $V = 900$ V.
2. Formula: $$V = k\frac{Q}{r} \Rightarrow r = k\frac{Q}{V}$$
3. Substitution: $$r = \frac{9 \times 10^9 \times 1 \times 10^{-6}}{900}$$
4. Calculation: $$r = \frac{9000}{900} = 10 \text{ m}$$
5. Final Answer: $$r = 10 \text{ m}$$

Example 4: Finding Charge from Given Potential

A point at a distance of 0.2 m from a charge has potential 4500 V. Find the value of the charge.

Solution:

1. Given: $V = 4500$ V, $r = 0.2$ m.
2. Formula: $$V = k\frac{Q}{r} \Rightarrow Q = \frac{Vr}{k}$$
3. Substitution: $$Q = \frac{4500 \times 0.2}{9 \times 10^9}$$
4. Calculation: $$Q = \frac{900}{9 \times 10^9} = 1 \times 10^{-7} \text{ C}$$
5. Final Answer: $$Q = 1 \times 10^{-7} \text{ C}$$

Example 5: Comparing Potentials at Two Distances

A point charge of $5 \mu C$ produces potentials $V_1$ and $V_2$ at distances 2 m and 5 m respectively. Find $V_1$, $V_2$, and the ratio $V_1:V_2$.

Solution:

1. Given: $Q = 5 \times 10^{-6}$ C, $r_1 = 2$ m, $r_2 = 5$ m.
2. Formula: $$V = k\frac{Q}{r}$$
3. Potential at 2 m: $$V_1 = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{2} = 22.5 \times 10^3 \text{ V}$$
4. Potential at 5 m: $$V_2 = 9 \times 10^9 \times \frac{5 \times 10^{-6}}{5} = 9 \times 10^3 \text{ V}$$
5. Ratio: $$\frac{V_1}{V_2} = \frac{22.5 \times 10^3}{9 \times 10^3} = \frac{5}{2}$$
6. Final Answer: $$V_1 = 2.25 \times 10^4 \text{ V}, \quad V_2 = 9 \times 10^3 \text{ V}, \quad V_1:V_2 = 5:2$$

Example 6: Effect of Changing Distance

The potential due to a point charge at a distance of 3 m is 600 V. What will be the potential at a distance of 9 m from the same charge?

Solution:

1. Since for a given charge, $$V \propto \frac{1}{r}$$
2. Therefore, $$\frac{V_1}{V_2} = \frac{r_2}{r_1}$$
3. Substitution: $$\frac{600}{V_2} = \frac{9}{3} = 3$$
4. So, $$V_2 = \frac{600}{3} = 200 \text{ V}$$
5. Final Answer: $$V_2 = 200 \text{ V}$$

Example 7: Potential Due to a Negative Charge at a Given Distance

Find the potential at a point 25 cm away from a charge of $-8 \times 10^{-8}$ C.

Solution:

1. Given: $Q = -8 \times 10^{-8}$ C, $r = 25$ cm $= 0.25$ m.
2. Formula: $$V = k\frac{Q}{r}$$
3. Substitution: $$V = 9 \times 10^9 \times \frac{-8 \times 10^{-8}}{0.25}$$
4. Calculation: $$V = 9 \times 10^9 \times -3.2 \times 10^{-7} = -28.8 \times 10^2 \text{ V}$$ $$V = -2880 \text{ V}$$
5. Final Answer: $$V = -2.88 \times 10^3 \text{ V}$$

Example 8: Distance When Potential Changes by a Factor

At a distance $r$ from a point charge, the potential is 120 V. At what distance will the potential become 30 V?

Solution:

1. For the same charge, $$V \propto \frac{1}{r}$$
2. Therefore, $$\frac{V_1}{V_2} = \frac{r_2}{r_1}$$
3. Substitution: $$\frac{120}{30} = \frac{r_2}{r}$$
4. So, $$4 = \frac{r_2}{r} \Rightarrow r_2 = 4r$$
5. Final Answer: The required distance is $$4r$$.