Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.
A small positive point charge of 7 mC moves from a distance of 9 cm to a distance of 17 cm.
How much work is done by the electric field?
Find the E field due to the long wire (Use Gauss law) at some distance r.
Work= E(9)q-E(17)q
To find the work done by the electric field, we can use the equation:
W = -ΔU
where W is the work done, ΔU is the change in potential energy.
The potential energy of a point charge in an electric field is given by:
U = q * V
where U is the potential energy, q is the charge, and V is the electric potential.
In this case, the electric potential can be calculated using the formula for the electric field created by a charged line:
E = λ / (2πε₀r)
where E is the electric field, λ is the linear charge density, ε₀ is the permittivity of free space, and r is the distance from the charged line.
First, let's calculate the electric field at a distance of 9 cm and 17 cm from the charged line:
r1 = 9 cm = 0.09 m
r2 = 17 cm = 0.17 m
E1 = (-2.50 nC/m) / (2πε₀ * 0.09 m)
E2 = (-2.50 nC/m) / (2πε₀ * 0.17 m)
Next, we can calculate the electric potential at each distance:
V1 = E1 * r1
V2 = E2 * r2
Now, we can calculate the change in potential energy:
ΔU = q * (V2 - V1)
Plugging in the values, we get:
ΔU = (7 mC) * ((E2 * r2) - (E1 * r1))
Finally, to find the work done by the electric field, we multiply the change in potential energy by -1:
W = -ΔU = - (7 mC) * ((E2 * r2) - (E1 * r1))
Calculating this value will give us the work done by the electric field as the small positive charge moves from a distance of 9 cm to 17 cm.