What is the work done by the electric force to move a 1 c charge from a to b?

Question:

Solution

The formula for calculating the work done by the electric force is

Work done (W) = Quantity of Charge (Q) x Voltage (V)

or W = QV

Hence, since the voltage flow is between the distances a and b, we can say that the voltage difference is

Voltage difference, V = V_a – V_b

This means W = QV will now become

W = Q(V_a – V_b)

Additionally, the quantity of charge is 1 c = 1 coloumb, which implies that Q = 1 c

Thus, W = Q(V_a – V_b) will now become

W = 1 x (V_a – V_b)

Hence,

W = V_a – V_b [where V_a is the voltage at a, and V_b is the voltage at b]

Therefore, the work done is the potential difference between points a and b.

Thus, the statement above is the answer to the question: “What is the work done by the electric force to move a 1 c charge from a to b?”

Explanation:

When electricity flows from point A to point B, we can now say that work is being carried out (work is done). Thus, a potential difference is the quantity of work done in moving a positive charge from a lower potential to a level of higher potential.

Potential difference (voltage), V = work done in moving a charge (W) / quantity of charge (Q)

Which implies that

V = W/Q

After cross multiplying the above formula to obtain work done

W = QV

Quantity of charge is measured in coloumbs, work is in joules, while potential difference is in volts.

We can also apply the following formulae to find work done in a circuit:

W=QV
W=IVt [Because Q=It, and t=time]
W=I²Rt [Because V=IR, and R=resistance in the circuit]
W=VQ x d [V = potential difference, and d=distance]

Examples of How to Find a Work Done in a Circuit

Here are a few examples with their solutions to guide you on how to find work done in a circuit:

Example 1

A 50μC charge flows through a conductor. If the potential difference between its ends is 20mV. Find the work done by the charge.

Solution

Data

Quantity of charge, Q = 50μC = 50 x 10^-6C

Potential difference, V = 20mV = 20 x 10^-3V

Work done, W =?

And W = QV

Which implies that

W = 50 x 10^-6 x 20 x 10^-3 = 10^-6 Joules or 1μJ

Example 2

A point charge of magnitude 2 x 10^-6 C is moved through a distance of 0.50m against a uniform electric field intensity of 30v/m. Determine the work done on the charge.

Solution

Data

Quantity of charge, Q = 2 x 10^-6 C

Electric field intensity = 30 v/m

Distance, d = 0.50 m

and the formula for electric field intensity is

E = F/Q

After cross multiplication Force, F = EQ

Work done, W = Force (F) x Distance (d)

Therefore, W = F x d = EQ x d = 30 x 2 x 10^-6 x 0.5 = 3 x 10^-5 Joules

Therefore, the work done is 3 x 10^-5 Joules

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