Question

QUESTION 3 Which of the following is true? (hint: use the Problem Solving insights offered in the book) a. A negative charge will always move from the outside of an equipotential surface to the inside. b. A positive charge will always move from a higher potential point to a lower potential point. c. A neutral charge will always move from a higher potential point to a lower potential point. d. A positive charge will always move from the inside of an equipotential surface to the outside. e. A negative charge will always move from a higher potential point to a lower potential point. ООООО QUESTION 4 A particle with a charge of -1.5 C and a mass of 2.5 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 42 m/s. What is the potential difference VB.V A between A and B? O a. -1.5x 103 V b. +1.5x 103 V 6.-2.1 x 10 3 v d. + 2.1 x 103 V e. 0.9x103 V

Answer 1

3) Any particle always tends to move from a point of higher potential energy to a point of lower potential energy.

Potential energy difference for a charge is given by [where V is the potential difference]

When a charged particle is released in an electric field, the work done by the electric force on the particle is always positive.

The relation between work done by an electric field and potential energy difference is

Therefore, we can see if V is negative and q is positive, the term -qV is positive, it means a positively charged particle always moves from higher potential to lower potential.

also, if V is positive, then q has to be negative for the work done to be positive, and thus we can say the negatively charged particle always moves from low potential to higher potential.

Therefore, the correct option is b. [answer]

4) Given:

q = -1.5 micro coulomb

m =

2.5 * 10-kg

v = 42 m/s

$V_{B}-V_{A}=?$

From Work energy principle, we can say

work done by electric force = change in kinetic energy

$W_{e}=\frac{1}{2}mv^{2}-0$

=> $-q(V_{B}-V_{A})=\frac{1}{2}mv^{2}-0$

$\Rightarrow (V_{B}-V_{A})=\frac{\frac{1}{2}mv^{2}}{-q}$

$=\frac{\frac{1}{2}*2.5*10^{-6}*42^{2}}{-(-1.5*10^{-6})}=1470V = 1.470*10^{3}V\approx+1.5*10^{3}V$

Therefore, the correct option is b. [answer]