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) \( 8 q_2 \)
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: