Question

When a particle of charge 10microcoulomb brought from infinity to point p 2 my of work is done by the external force what is the potential at p

Answer 1

Answer:

Potential at a point in a field is defined as the amount of work done in bringing a unit positive test charge from infinity to that point along any arbitrary path. mathematically;

$\footnotesize\longrightarrow\:\sf V = \dfrac{W}{q_{0}}$

• Work done = 2 mJ
• Test charge = 10 μC

$\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times {10}^{ - 3} }{10 \times {10}^{ - 6} }$

$\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times {10}^{ - 3} \times {10}^{6} }{10 }$

$\footnotesize\longrightarrow\:\sf V = \dfrac{2 \times {10}^{3} }{10 }$

$\footnotesize\longrightarrow\:\sf V = 0.2\times {10}^{3}$

$\footnotesize\longrightarrow\:\sf V = 2\times {10}^{ - 1} \times {10}^{3}$

$\footnotesize\longrightarrow\:\sf V = 2 \times {10}^{2}$

$\footnotesize\longrightarrow\: \underline{ \underline{\sf V = 2 00 \: J/C}}$

• MORE INFORMATION:

$\footnotesize\maltese \: \:\sf 1 \: mC = {10}^{ - 3} C$

$\footnotesize\maltese \: \:\sf 1 \: \mu C = {10}^{ - 6} C$

$\footnotesize\maltese \: \:\sf 1 \: n C = {10}^{ - 9} C$

$\footnotesize\maltese \: \:\sf 1 \: mJ = {10}^{ - 3} J$

• Electric Potential is a scalar quantity.

• S.I unit is Joule/Coulomb (J/C) or Volt (v)

• Dimensional Formula = [V] = [ML²T-³A-¹]

Answer 2

Answer:

In the question the unit of the work done,is missing.

I assume that it is 2 mJ .

Potential at a specific point V = W/Q [ where V is in volt when W is in Jule and Q is in coulomb]

Here, W = 2mJ = 2 × 10-³ J

Q = 10 microcoulmb= 10 × 10-⁶ C

Therefore, the potential difference at p = 2× 10-³/10×10-⁶ V = 200 V