Answer:
[tex]\Delta V = 1240 Volts[/tex]
Explanation:
As we know that charge moves from higher potential to lower potential
so here change in electrostatic potential energy must be equal to final kinetic energy of the charge
so we will have
[tex]q\Delta V = \frac{1}{2}mv^2[/tex]
so we will have
[tex]q = 8\times 10^{-8} C[/tex]
[tex]m = 5.5 \times 10^{-10} kg[/tex]
[tex]v_f = 600 m/s[/tex]
[tex](8 \times 10^{-8})\Delta V = \frac{1}{2}(5.5 \times 10^{-10})(600^2)[/tex]
[tex]\Delta V = 1240 Volts[/tex]
Answer:
B. 1240 V
Explanation:
Given
Initial velocity of the charge, u = 0 m/s
Final velocity of the charge, v = 600 m/s
Magnitude of the charge, q = [tex]8.0 \times 10^{-8}C[/tex]
mass of the charge, m = [tex]5.5 \times 10^{-10} kg[/tex]
Solution
Gain in mechanical energy = Loss in electric potential energy
[tex]\frac{1}{2} mv^{2} - \frac{1}{2} mu^{2}= Vq\\\\0.5 \times 5.5 \times 10^{-10} \times 600^{2} - 0 = V \times 8.0 \times 10^{-8}\\\\V = 1237.5 V[/tex]
Rounding off the answer to closest tens, we get
V = 1240 V
