Question

The work done in moving a -5μC charge in an electric field $ \overrightarrow{E} = (8r sin \theta \hat{r}+ + 4r cos \theta \hat{\theta} ) V/m$, from a point $A(r, \theta) = (10,\frac{\pi}{6})$ to a point $B(r,\theta) (10.\frac{\pi}{6})$ is _____ $mj$.

Answer 1

Correct Answer: -1

Calculate Potential at B ($V_B$) for $(10, \frac{\pi}{2})$:

$$V_B = -4(10)^2 \sin\left(\frac{\pi}{2}\right)$$

$$V_B = -4(100) (1) = -400 \, \text{V}$$

Calculate Work Done (for the likely intended case):

$$W = q(V_B - V_A)$$

$$W = (-5 \times 10^{-6}) \times [-400 - (-200)]$$

$$W = (-5 \times 10^{-6}) \times [-200]$$

$$W = 1000 \times 10^{-6} \, \text{J}$$

$$W = 1 \, \text{mJ}$$

Answer

The work done is 1 mJ.