A,B,C and D are four points on an imaginary circle in region containing uniform electric field as shown in figure. Select the incorrect option
Electric potential decreases in the direction of electric field. so , \[V_{B} < V_{D}\] is wrong.
An electric charge 10–³ μC is placed at the origin (0, 0) of X–Y co-ordinate system. Two points A and B are situated at ( √2, √2) and (2, 0) respectively. Find the potential difference between the points A and B.
The potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by:
\[
V = \frac{kq}{r}
\]
1. Distance from the charge at the origin:
- Point \( A(\sqrt{2}, \sqrt{2}) \): Distance \( r_A = \sqrt{(\sqrt{2})^2 + (\sqrt{2})^2} = \sqrt{4} = 2 \).
- Point \( B(2, 0) \): Distance \( r_B = \sqrt{(2)^2 + 0^2} = 2 \).
2. Potentials at \( A \) and \( B \):
\[
V_A = \frac{kq}{r_A} = \frac{kq}{2}, \quad V_B = \frac{kq}{r_B} = \frac{kq}{2}.
\]
3. Potential difference:
\[
V_A - V_B = \frac{kq}{2} - \frac{kq}{2} = 0.
\]
Thus, \( V_A - V_B = 0 \).
Find final charge on smaller sphere after closing switch :
After closing the switch:
1. Potential equality: \(\frac{Q_1}{R} = \frac{Q_2}{3R} \Rightarrow Q_2 = 3Q_1\).
2. Charge conservation: \(Q_1 + Q_2 = 36Q \Rightarrow Q_1 + 3Q_1 = 36Q \Rightarrow Q_1 = 9Q\).
Final charge on the smaller sphere = 9Q.
There are two concentric conducting shells. The potential of outer shell is 10 V and that of inner shell is 15 V. If the outer shell is grounded, the potential of inner shell becomes/remains
When the outer shell is grounded, its potential becomes 0. The potential difference between the shells remains the same as before grounding (since grounding doesn't change the relative configuration of charges).
Initial potential difference:
Inner shell - Outer shell = \( 15 \, \text{V} - 10 \, \text{V} = 5 \, \text{V} \).
After grounding, the inner shell’s potential relative to the grounded outer shell becomes \( 0 + 5 \, \text{V} = 5 \, \text{V} \).
Thus, the inner shell’s potential is 5 V.
Figure shows a charged conductor of irregular shape. If \ ( \sigma_{A},\sigma_{B} and \sigma_{C}\) are the surface charge densities at A, B and C respectively, then :
Pointed ends have more surface charge density so,
\[\sigma_{A}>\sigma_{B} > \sigma_{C}\]