How much work is done by the electric force when a 0.50-C charge is moved from the point (3.0 cm, 2.0 cm) to the point (8.0 cm, 4.0 cm)

This question is incomplete, the complete question is;

How much work is done by the electric force when a 0.50-C charge is moved from the point (3.0 cm, 2.0 cm) to the point (8.0 cm, 4.0 cm)? (the electric potential at 3cm, 2cm is 20V), (the electric potential at 8cm, 4cm is 30V)

work is done by the electric force is 5J

Explanation:

Given that;

Charge C = 0.5 C

Electric potential Ep1 = 20 V

Electric Potential Ep2 = 30 V

work is done by the electric force = ?

now to calculate work is done by the electric force, we say

work is done by the electric force = charge × ( potential drop)

work is done by the electric force = C × ( Ep2 - Ep1 0

work is done by the electric force =  0.50 × ( 30 - 20)

work is done by the electric force = 5J

  the unit of the measure =

explanation:

  the property that is quantified by the energy needed per of material to raise the temperature by one.

  si unit of energy = joule (j)

  si unit of temperature = kelvin (k)

  si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.