How much work would be done by the electric field if $Q = 20 C$ were moved from $A (8 V)$ to $B (3.5 V)$ ?

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Solution

Step 1: Given
Charge $Q = 20 C$
Voltage at point $A, V_{A} = 8 V$
Voltage at point $B, V_{B} = 3.5 V$
Step 2: Calculate the potential difference
We know that potential difference is the amount of work required to bring a unit positive charge from one point to another.
Hence, the potential difference
$V = V_{B} - V_{A}$
$= - 3.5 - 8$
$V = - 4.5 V$
Step 3: Calculate the work done
We know that work done by the electric field opposes the work done by the external agent
$W = - Q V$
Upon substituting the values we get,
$W = - \{20 \times (- 4.5)\}$
$= 90 J$
Hence, $90 J$ work is done by the electric field.

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