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Question: A ball of mass \(1g\) and charge \({10^{ - 8}}C\) moves from A, where potential is \(600V\) , to B, ...

A ball of mass 1g1g and charge 10−8C{10^{ - 8}}C moves from A, where potential is 600V600V , to B, where potential is 0. The velocity of the ball at B is 20cm/s20cm/s . The velocity of the ball at point A must have been:
(A) 16.8cm/s16.8cm/s
(B) 22.8cm/s22.8cm/s
(C) 228cm/s228cm/s
(D) 168cm/s168cm/s

Explanation

Solution

Energy is a property that can be transferred to or from an object in order to perform work. Energy can only be transferred from one state to another but cannot be created or destroyed. For example, in an object at rest, it possesses potential energy while when the body is in motion it possesses kinetic energy.
Formulas used: We will be using the formula to find the kinetic energy of a body, KE=12mv2KE = \dfrac{1}{2}m{v^2} where KEKE is the kinetic energy of the body, mm is the mass of the body, and vv is the kinetic energy of the body.
We will also be using the formula to find the electric energy, Eelec=Q×V{E_{elec}} = Q \times V where Eelec{E_{elec}} is the electrical energy of the body, QQ is the charge carried by the body, and VV is the electric potential of the charge carrying body.

Complete Step by Step answer:
We know that energy of a body can either be transformed or stored in them, but can never be created or destroyed. Thus, the energy in any system at any point of time is always conserved.
From the problem we can infer that the ball has a mass of m=1g=10−3kgm = 1g = {10^{ - 3}}kg and the charge on the ball is Q=10−8CQ = {10^{ - 8}}C . We also know that the ball is moving from point A to point B, where the potential at point A is VA=600V{V_A} = 600V while the electric potential and point B is VB=0V{V_B} = 0V . We are also given the velocity of the ball at point B which is vB=20cm/s=0.2cm/s{v_B} = 20cm/s = 0.2cm/s and we are required to find the velocity of the ball at point A.
Since the energy of the system needs to be conserved, we know that the change in kinetic energy of the ball will be equal to the change in electric potential energy of the ball due to the charges on it.
Change in Kinetic energy of the ball is given by, ΔKE=12mvB2−12mvA2\Delta KE = \dfrac{1}{2}m{v_B}^2 - \dfrac{1}{2}mv_A^2 and the change in electric potential energy of the ball is given by, ΔEelec=(Q×VB)−(Q×VA)\Delta {E_{elec}} = \left( {Q \times {V_B}} \right) - \left( {Q \times {V_A}} \right) .
According to the law of conservation of energy,
12mvB2−12mvA2=(Q×VB)−(Q×VA)\dfrac{1}{2}m{v_B}^2 - \dfrac{1}{2}mv_A^2 = \left( {Q \times {V_B}} \right) - \left( {Q \times {V_A}} \right)
12m(vB2−vA2)=Q(VB−VA)\dfrac{1}{2}m\left( {{v_B}^2 - v_A^2} \right) = Q\left( {{V_B} - {V_A}} \right)
Substituting the known values, we get
12×10−3×((0.2)2−vA2)=10−8×(0−600)\dfrac{1}{2} \times {10^{ - 3}} \times \left( {{{\left( {0.2} \right)}^2} - v_A^2} \right) = {10^{ - 8}} \times \left( {0 - 600} \right)
(0.2)2−vA2=10−8×600×210−3{\left( {0.2} \right)^2} - v_A^2 = \dfrac{{{{10}^{ - 8}} \times 600 \times 2}}{{{{10}^{ - 3}}}}
Simplifying the equation, we get,
(0.2)2−vA2=12×10−3{\left( {0.2} \right)^2} - v_A^2 = 12 \times {10^{ - 3}}
vA2=12×10−3+40×10−3v_A^2 = 12 \times {10^{ - 3}} + 40 \times {10^{ - 3}}
Now solving and applying square root on both sides we get,
vA=52×10−3{v_A} = \sqrt {52 \times {{10}^{ - 3}}}
⇒vA=0.228m/s\Rightarrow {v_A} = 0.228m/s
Thus, the velocity at point A will be vA=22.8cm/s{v_A} = 22.8cm/s .

Hence the correct answer will be option B.

Note: We can see that the kinetic energy of the body is conserved as the electrical potential energy in this system. If considered in a magnetic energy system, the energy will still be conserved.