To determine how much work the electric field does in moving a charge, we use the formula for electrical work done:
[tex]\[ W = q \Delta V \][/tex]
Here's the step-by-step solution:
1. Identify the given values:
- The charge [tex]\( q \)[/tex] is [tex]\( -7.3 \)[/tex] microcoulombs ([tex]\(\mu C\)[/tex]). Since 1 microcoulomb ([tex]\( \mu C \)[/tex]) is [tex]\( 1 \times 10^{-6} \)[/tex] coulombs, we convert the charge to coulombs:
[tex]\[ q = -7.3 \times 10^{-6} \text{ C} \][/tex]
- The potential difference [tex]\( \Delta V \)[/tex] is [tex]\( +65 \)[/tex] volts (V).
2. Substitute the given values into the formula:
[tex]\[ W = q \Delta V \][/tex]
[tex]\[ W = (-7.3 \times 10^{-6} \text{ C}) \times (65 \text{ V}) \][/tex]
3. Calculate the work done:
The product of the charge and the potential difference gives:
[tex]\[ W = -0.00047450000000000004 \text{ joules} \][/tex]
4. Express the answer with two significant figures:
Since we need the result to two significant figures, we round the answer appropriately:
[tex]\[ W \approx -4.7 \times 10^{-4} \text{ J} \][/tex]
So the electric field does approximately [tex]\( -4.7 \times 10^{-4} \)[/tex] joules of work in moving a [tex]\( -7.3 \mu C \)[/tex] charge from ground to a point whose potential is [tex]\( +65 V \)[/tex] higher.
