In figure, a particle having mass m=5 ~g and charge q^ =2 10⁻⁹ C starts from rest at point a and moves in a straight line to point b . What is its speed v at point b ?

Step-by-step solution

According to conservation of energy, we get K_a+U_a=K_b+U_b Here, K_a=0 and the potential energies are U_a=q^ V_a and U_b=q^ V_b aligned & 0+q^ V_a= 1 2 m v^2+q^ V_b & or v= 2 q^ (V_a-V_b ) m & V_a= (9.0 10^9 Nm ^2 C ⁻² ) aligned array r ( 3 10⁻⁹ C 0.01 ~m + -3 10⁻⁹ C 0.02 ~m )=1350 ~V V_b= (9.0 10^9 Nm ^2 C ⁻² ) ( 3 10⁻⁹ C 0.02 ~m + -3 10⁻⁹ C 0.01 ~m )=-1350 ~V . array aligned & v= 2 (2 10⁻⁹ C )(2700 ~V ) 5 10⁻³ ~kg =4.65 10⁻² ~m ~s ⁻¹ & =4.65 ~cm ~s ⁻¹ aligned