Question

A particle with a charge of -4.4 μC and a mass of 3.5 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 26 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VBis less than VA, then give the answer as a negative number.

help me please

Answer 1

since charge on the particle = - 4.4 $\mu$ C = q (let)

mass of the particle = 3.5 * 10^-6 kg

and speed of the particle at the point B = 26m/s

since, total energy in the electric field is conserve.

so, from the conservation of energy we know that

change in kinetic energy = negative of the change in potentional energy.

   $\Delta$ K = - $\Delta$ U

initially, particle is in rest so kinetic energy at that point is zero, and kinetic energy at point B is 1/2mv²

so, $\Delta$ K = 1/2mv²

and electric potential energy of the charge particle = q $\Delta$ V = $\Delta$ U

where q = charge and $\Delta$ V = potential difference between two points A and B

now from the above equation,

1/2mv² = -q $\Delta$ V

$\Delta$ V = (1/2mv²)/q

  = -1/2*3.5*10^-6 * (26 )²*/- 4.4 =+2.69 * 10^-6 volt

$\Delta$ V= +2.69 * 10^-6 volt

now if (a)V_BV_A then VB - V_A 0 that gives the positive no. because any no. that is greater than zero is positive.

and again, if(b) V_B V_{A ,} V_B - V_A0, that gives the negative no. because the no. less than zero is negative

from the condition (a) we find that potential difference between A and B will be positive +2.69 * 10^-6 volt

and for the condition (b) potential difference will be negative -2.69 * 10^-6 volt