Question

The potential at a point due to charge of 5×10-7C located 10cm away is​

Answer 1

Given:

Charge = 5 × 10⁻⁷ C

The distance between the charge and the point = 10 cm = $\frac{10}{100}$ m = 0.1 m

To find:

Electric Potential at the point

Solution:

We know that at any point, the electric potential at a point due to a charge separated by a distance is given by,

$\boxed{\bold{V = k \frac{q}{r} }}$

where

V = electric potential

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

q = charge

r = distance of separation

Now, we will substitute the given values of q and r in the formula above,

V = 9 × 10⁹  × $\frac{5\:\times\:10^-^7\:C}{0.1\:m}$

⇒ V = 450 × 10⁹ × 10⁻⁷

⇒ V = 450 × 10⁹⁻⁷

⇒ V = 450 × 10²

⇒ V = 45000 × $\frac{10^4}{10^-^4}$

⇒ V = 4.5 × 10⁴ volts

Thus, the electric potential at a point due to charge of 5 × 10⁻⁷ C located 10 cm away is​ 4.5 × 10⁴ volts.

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Answer 2

Given:

Distance, r = 10 cm

On converting "cm" into "m",

r = $\frac{10}{100}$

 = $0.1 \ m$

Charge, q = $5\times 10^{-7} \ C$

We know the value of k,

Coulomb's constant, k = $9\times 10^{9} \ N-m^2/C^2$

To Find:

Electrical potential, V = ?

Solution:

As we know,

$V=k\times \frac{q}{r}$

On putting the given values in the above expression, we get

⇒    $=9\times 10^9\times (\frac{5\times 10^{-7} }{0.1} )$

⇒    $=450\times 10^{9}\times 10^{-7}$

When the base are same then powers will add,

⇒    $=450\times 10^{9-7}$

⇒    $=450\times 10^2$

⇒    $=45000\times \frac{10^4}{10^{-4}}$

⇒    $=4.5\times 10^4 \ volts$

So that the electrical potential at that point will be "4.5 × 10⁴ volts".