Given: A graph of electric potential \( V \) along the X-axis. We are to choose the correct graph of
electric field strength \( E \). Key Concept: Electric field is the negative gradient (slope) of electric potential: \[ E = -\frac{dV}{dx} \] - Where the potential decreases linearly, \( E \) is constant and negative. - Where the slope of \( V(x) \) increases, \( E \) becomes more negative. - Where the slope becomes less steep (towards flat), \( E \) approaches zero. - If slope changes sign, \( E \) changes sign.
Analyzing the given potential graph: - From left to right, the slope (negative of \( E \)) is: - Steep (large negative) - Moderate (less negative) - Flat (zero) - Positive slope (negative \( E \)), so field becomes positive So, \( E \) starts negative, increases toward zero, then becomes positive.
Correct Plot: The electric field graph that starts negative, increases, crosses zero, and becomes positive.
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?
A quantity \( X \) is given by: \[ X = \frac{\epsilon_0 L \Delta V}{\Delta t} \] where:
- \( \epsilon_0 \) is the permittivity of free space,
- \( L \) is the length,
- \( \Delta V \) is the potential difference,
- \( \Delta t \) is the time interval.
The dimension of \( X \) is the same as that of: