Two charges $ q_1 $ and $ q_2 $ are separated by a distance of 30 cm. A third charge $ q_3 $ initially at C as shown in the figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to the movement of $ q_3 $ from C to D is given by $ \frac{q_3 K}{4 \pi \epsilon_0} $, the value of $ K $ is:
We are given that: The charge \( q_1 \) and \( q_2 \) are separated by 30 cm. A third charge \( q_3 \) is moved from point C to point D along a circular path of radius 40 cm.
The change in potential energy is given by \( \frac{q_3 K}{4 \pi \epsilon_0} \).
The potential energy of a system of charges is given by: \[ U = \frac{q_1 q_2}{4 \pi \epsilon_0 r} \]
Where \( r \) is the distance between the charges. In this case, the distance changes as \( q_3 \) moves from C to D.
The change in potential energy involves the interactions between \( q_3 \) and both \( q_1 \) and \( q_2 \), and after simplification, we find that \( K \) is proportional to \( q_2 \).
Final Answer (1) \( 8 q_2 \)
A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $ \epsilon_1 $ and $ \epsilon_2 $, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $ C_1 $ and $ C_2 $ respectively, then $ \frac{C_1}{C_2} $ is:
Three parallel plate capacitors $ C_1 $, $ C_2 $, and $ C_3 $ each of capacitance 5 µF are connected as shown in the figure. The effective capacitance between points A and B, when the space between the parallel plates of $ C_1 $ capacitor is filled with a dielectric medium having dielectric constant of 4, is: