Question 785: Electrostatics & Capacitance

Question

The work done in moving a charge particle of charge $2 	imes 10^{-8} \mathrm{C}$ between two points having potential difference of 36 V is

Accepted Answer

To calculate the work done in moving a charged particle with a charge of $2 	imes 10^{-8} \, 	ext{C}$ across a potential difference of 36 V, we use the formula:
$  W = q \cdot \Delta V  $
Where:
$ W $ is the work done,
$ q = 2 	imes 10^{-8} \, 	ext{C} $ (charge of the particle),
$ \Delta V = 36 \, 	ext{V} $ (potential difference).
Substituting the given values into the equation, we get:
$  W = 2 	imes 10^{-8} 	imes 36  $
$  W = 72 	imes 10^{-8} \, 	ext{J}  $
Converting the result to microjoules ($\mu J$):
$  W = 0.72 	imes 10^{-6} \, 	ext{J}  $
$  W = 0.72 \, \mu 	ext{J}  $
The work done in moving the charge is $0.72 \, \mu 	ext{J}$.

Answer

$1.8 \mu \mathrm{~V}$

Answer

$0.72 \mu \mathrm{~V}$

Answer

$3.6 \mu \mathrm{~V}$

Answer

$1.44 \mu \mathrm{~V}$

Electrostatics & Capacitance

The work done in moving a charge particle of charge 2 \times 10^{-8} \mathrm{C} between two points having potential difference of 36 V is

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