Question

Chapter 19, Problem 04 GO A particle has a charge o +2.8 μС and moves from point A to pont B a distance of 0 17 m. The particle experiences a costant electrc force, and ts mer n of the force. The difference between the particle's electric potential energy at A and B is EPEA EPE+8.0 x 10 J. (a) Find the magnitude of the electric force that acts on the ng the loco acon particle. (b) Find the magnitude of the electric field that the particle experiences (a) Number (b) Number Units Units

Answer 1

Problem # 4

Part (a) - Solution

$\:$

The magnitude of the electric force is related to the electric potential energy by the eq.

$\:$

$W=\left | \Delta U \right |=F_{e}\Delta x$

$\Rightarrow \: \: \: F_{e} =\frac{\left | \Delta U \right |}{\Delta x}$

$\Rightarrow \: \: \: F_{e} =\frac{\left | 8.0\times 10^{-4} \, J \right |}{0.17m}$

$\:$

or

$\:$

${\color{Blue} F_{e} =4.7\times 10^{-3} \, N}$

$\:$

Part (b) - Solution

$\:$

The magnitude of the electric field is related to the electric force by the equation

$\:$

$E=\frac{F_{e}}{q}$

$\Rightarrow \: \: \: E=\frac{4.7\times 10^{-3} \, N}{2.8\times 10^{-6} \, C}$

$\:$

or

$\:$

${\color{Blue} E=1.68\times 10^{3} \, N/C}$

$\:$

Problem # 2

$\:$

Using Principle of conservation of energy,

$\:$

$E_{A}=E_{B}$

$\Rightarrow \: \: \: K_{A}+U_{A}=K_{B}+U_{B}$

$\Rightarrow \: \: \: \frac{mv_{A}^{2}}{2}+qV_{A}=\frac{mv_{B}^{2}}{2}+qV_{B}$

$\Rightarrow \: \: \: qV_{B}-qV_{A}=\frac{mv_{A}^{2}}{2}+\frac{mv_{B}^{2}}{2}$

$\Rightarrow \: \: \: V_{B}-V_{A}=\frac{m(v_{A}^{2}-v_{B}^{2})}{2q}$

$\Rightarrow \: \: \: V_{B}-V_{A}=\frac{(2.5\times 10^{-6} \, kg)((0)^{2}-(43\, m/s)^{2})}{2(-2.8\times 10^{-6}\, C)}$

$\:$

or

$\:$

${\color{Blue} V_{B}-V_{A}=825 \, Volts}$