Question

A charge q of 7.3 × 10^15 coulomb moves from point A to point B in an electric field. If the potential energy of the charge at point A is 3.5 × 10^-12 joules and that at point B is 1.3 × 10^-12 joules, what is the potential difference between point A and B?
A. 1.3 × 10^2 volts
B. 3.0 × 10^2 volts
C. 3.5 × 10^3 volts
D. 6.8 × 10^3 volts

Answer 1

The charge is 7.3 × $10^{- 15}$ C

If we call E the potential energy and q the charge then we know that the potential (V) is:

$V = \frac{E}{q}$

So:

$V_A = \frac{3.5\ x\ 10^{-12}}{7.3\ x\ 10^{-15}}\\\\V_A = 479.45\ Volts\\\\V_B = \frac{1.3\ x\ 10 ^{-12}}{7.3\x\ 10 ^{-15}}\\\\V_B = 178.08\ Volts.$

Finally:

$V_A - V_B = 479.45 - 178.08$

$V_A - V_B = 301\ Volts\\\\V= 3.0\ x\ 10^2$

The answer is option B.

Answer 2

Answer  

Option B - 3.0 × 10^2 volts  

Step by step explanation:  

Given in the question:

q= of 7.3 × 10^-15 coulombs

EA =3.5 × 10^-12 joules  

EB= 1.3 × 10^-12 joules

As per the required:

△v= ?

By using the formula and substituting the given in the equation,

E= q △v

To find △v:  

△v = E / q  

△v= (3.5 × 10^-12 -  1.3 × 10^-12) /  7.3 × 10^-15

△v = 3.0 × 10^2 volts.