Question

Calculate the potential at a point having coordinates (0.3 m, 0.4 m, 0.5m) due to a charge of 20 μC placed at the origin.

Answer 1

$\mathtt{ \huge{\fbox{Solution :)}}}$

Gievn ,

• The charge is 20 μC or 20 × (10)^-6 C

Let , the point P whose vertices is (0.3 m, 0.4 m, 0.5m)

So , the distance of a point P (0.3 , 0.4 , 0.5 ) m from origin (0 , 0 , 0) m

We know that ,

$\large{ \mathtt{ \fbox{D = \sqrt{ {( x_{2} - x_{1}) }^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2} } \: \: \: }}}$

Thus ,

$\sf \mapsto D = \sqrt{ {(0 - 0.3)}^{2} + {(0 -0.4)}^{2} + {(0 - 05)}^{2} } \\ \\\sf \mapsto D = \sqrt{0.09 + 0.16 + 0.25 } \\ \\\sf \mapsto D = \sqrt{0.5} \: \: m$

We know that , potential difference due to point charge q at distance r from it is given by

$\mathtt{ \large{ \fbox{V = \frac{1}{4\pi e_{0} } \frac{q}{r} }}}$

Thus ,

$\sf \mapsto V = \frac{9 \times {(10)}^{9} \times 20 \times {(10)}^{ - 6} }{ \sqrt{0.5} } \\ \\ \sf \mapsto V = \frac{180 \times {(10)}^{3} }{ \sqrt{0.5} } \\ \\ \sf \mapsto V = \frac{0.5 \times 180 \times {(10)}^{4} }{5} \\ \\ \sf \mapsto V =\sqrt{0.5} \times 36 \times {(10)}^{4} \: \: \: volt$

Hence , the potential difference is $\mathtt{ \fbox{\sqrt{0.5} \times 36 \times {(10)}^{4} }}$volt

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Answer 2

The electric potential at that point is 2.5 × 10⁴ volts.

Explanation:

The distance between given point (0.3 m, 0.4 m, 0.5 m) and origin is given as:

r = √((0 - 0.3)² + (0 - 0.4)² + (0 - 0.5)²)

r = √(0.09 + 0.16 + 0.25)

r = √(0.5)

∴ r = 0.707 m

The electric potential is given by the formula:

V = kq/r

Where,

k = Coulomb's constant = 9 × 10⁹ N.m²/C²

q = Charge = 20 μC = 20 × 10⁻⁶ C

r = Distance between the points = 0.707 m

On substituting the values, we get,

V = (9 × 10⁹ × 20 × 10⁻⁶)/0.707

V = (18 × 10⁴)/0.707

∴ V = 2.5 × 10⁴ volts