Question

An object with charge q = −5.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. After the charge has moved to point B, 0.500 m to the right, it has kinetic energy 4.00×10−7 J .

Part A

If the electric potential at point A is +30.0 V, what is the electric potential at point B?
Part B
What is the magnitude of the electric field?
Part C
What is the direction of the electric field?

from point A to point B
perpendicular to the line AB
from point B to point A

Answer 1

Answer

given,

q = −5.00 × 10⁻⁹ C

distance, r = 0.5 m

K.E = 4.00×10⁻⁷J

a) V_a = 30 V

   Using law of conservation of energy

    KE = (V_a-V_b)q

  $V_b = V_a + \dfrac{KE}{q}$

  $V_b = 30 + \dfrac{4\times 10^{-7}}{5\times 10^{-9}}$

     V_b = 110 V

b) Electric field calculation

  $E = \dfrac{V_b-V_a}{d}$

  $E = \dfrac{110-30}{0.5}$

      E = 160 V/m

c) The direction of electric field is from point B to point A because electric potential of point B is greater than Point A.

Answer 2

This question involves the concepts of the law of conservation of energy, kinetic energy, electric potential energy, and electric field.

A) The electric potential at point B is "110 volts".

B) The magnitude of the electric field is "100 volt/m".

C) The direction of the electric field is "from point B to point A".

A)

According to the law of conservation of energy the kinetic energy of the object must be equal to the electric potential energy lost:

$K.E = (\Delta V)(q)\\\\\frac{K.E}{q} = V_A-V_B$

where,

K.E = kinetic energy = 4 x 10⁻⁷ J

q = charge = - 5 x 10⁻⁹ C

$V_A$ = electric potential at A = 30 V

$V_B$ = Electric potential at B = ?

Therefore,

$\frac{4\ x\ 10^{-7}\ J}{-5\ x\ 10^{-9}\ C}=30\ V-V_B\\\\V_B=30\ V+80\ V\\\\V_B=110\ volts$

B)

The magnitude of the electric field is given by the following formula:

$E=\frac{\Delta V}{d}$

where,

E = electric field magnitude = ?

ΔV = change in electric potential = 80 V - 30 V = 50 V

d = distance between points = 0.5 m

Therefore,

$E=\frac{50\ V}{0.5\ m}\\\\E=100\ volt/m$

C)

The direction of the electric field is always from high potential to low potential. Hence, the direction will be from point B to point A.

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.