Two electric charges 9 macro c and -3 macro c placed 0.16apart in air ...
Two electric charges 9 macro c and -3 macro c placed 0.16apart in air ...
Understanding Electric Potential
Electric potential (V) at a point due to a charge is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where:
- \( V \) = electric potential,
- \( k \) = Coulomb's constant (\( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
- \( Q \) = charge,
- \( r \) = distance from the charge to the point.
Given charges:
- \( Q_1 = 9 \, \mu C = 9 \times 10^{-6} \, C \) (positive charge)
- \( Q_2 = -3 \, \mu C = -3 \times 10^{-6} \, C \) (negative charge)
- Distance between charges = 0.16 m
Calculating Potential at Point A
- **Location**: 0.04 m from \( Q_2 \) (negative charge) and 0.12 m from \( Q_1 \) (positive charge).
- **Potential at A**:
\[
V_A = V_{Q_1} + V_{Q_2}
\]
\[
V_{Q_1} = \frac{8.99 \times 10^9 \times 9 \times 10^{-6}}{0.12} = 67425 \, V
\]
\[
V_{Q_2} = \frac{8.99 \times 10^9 \times (-3) \times 10^{-6}}{0.04} = -67425 \, V
\]
\[
V_A = 67425 - 67425 = 0 \, V
\]
Calculating Potential at Point B
- **Location**: 0.08 m from \( Q_2 \) (outside the charges).
- **Potential at B**:
\[
V_B = V_{Q_1} + V_{Q_2}
\]
\[
V_{Q_1} = \frac{8.99 \times 10^9 \times 9 \times 10^{-6}}{0.24} = 28142.5 \, V
\]
\[
V_{Q_2} = \frac{8.99 \times 10^9 \times (-3) \times 10^{-6}}{0.08} = -33737.5 \, V
\]
\[
V_B = 28142.5 - 33737.5 = -5595 \, V
\]
Summary of Results
- **Potential at A**: \( 0 \, V \)
- **Potential at B**: \( -5595 \, V \)
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