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Physics · 2024 · 3 marks
CBSE 2024 · Region 4 · Set 1 · Q23
The electric field in a region is given by E =(10 x+4) i where x is in m and E is in N/C. Calculate the amount of work done in taking a unit charge…
The electric field in a region is given by \[\overrightarrow{\mathrm{E}}=(10 x+4) \hat{\mathrm{i}} \] where $\displaystyle x$ is in m and E is in N/C. Calculate the amount of work done in taking a unit charge from
(i)
$\displaystyle (5 \mathrm{~m}, 0)$ to $\displaystyle (10 \mathrm{~m}, 0)$
(ii)
$\displaystyle (5 \mathrm{~m}, 0)$ to $\displaystyle (5 \mathrm{~m}, 10 \mathrm{~m})$
Marking-scheme solution
(i)
$\displaystyle \Delta V=-\int_{x_{1}}^{x_{2}} E d x$
$\displaystyle \Delta V=-\int_{5}^{10}(10 x+4) d x=-\left[\frac{10 x^{2}}{2}+4 x\right]_{5}^{10}$
$\displaystyle =-395 \mathrm{~V}$
$\displaystyle W=q \Delta V=-395 \times 1$
$\displaystyle =-395 \mathrm{~J}$
(ii)
$\displaystyle \Delta V=-\int_{x_{1}}^{x_{2}} E d x$
$\displaystyle \Delta V=-\int_{5}^{5}(10 x+4) d x=0$
$\displaystyle W=q \cdot \Delta V=0$
OR
Displacement is perpendicular to the electric field, therefore:
$\displaystyle \Delta V=0$
$\displaystyle W=q \cdot \Delta V=0$
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