Write two characteristics of equipotential surfaces. A uniform electric field of 50 NC\(^{-1}\) is set up in a region along the \( x \)-axis. If the potential at the origin \( (0, 0) \) is 220 V, find the potential at a point \( (4m, 3m) \).
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The potential in a uniform electric field is a linear function of distance along the direction of the field.
Constant Potential: The electric potential is the same at all points on an equipotential surface, meaning no work is done in moving a charge along the surface.
Perpendicular to Electric Field: Equipotential surfaces are always perpendicular to the electric field lines. This is because the electric field is defined as the gradient of the potential, so it must be normal (perpendicular) to the equipotential surfaces.
2. Given Information:
The electric field is uniform and directed along the \( x \)-axis.
The potential at any point is given by the equation: \[ V = V_0 - E \cdot d \] where:
\( V_0 \) is the potential at the origin,
\( E \) is the magnitude of the electric field,
\( d \) is the distance along the \( x \)-axis.
3. Calculation of Potential at \( (4m, 3m) \):
Since the electric field is along the \( x \)-axis, the distance \( d \) along the \( x \)-axis is 4 m (ignoring the \( y \)-coordinate of 3 m because the field is only in the \( x \)-direction).
The potential at this point is given by the formula: \[ V = 220 \, \text{V} - 50 \, \text{NC}^{-1} \cdot 4 \, \text{m} = 220 \, \text{V} - 200 \, \text{V} = 20 \, \text{V} \]
4. Conclusion:
Thus, the potential at the point \( (4m, 3m) \) is \( 20 \, \text{V} \).
