Question

12. At each corner of an equilateral triangle charges of
+250 C are placed in air. What will be the resultant
electric potential at the centroid of the triangle whose
distance from each corner is 18 cm? Ans. 3.75 x 107
​

Answer 1

Answer:

$\boxed{\mathfrak{Result \ electric \ potential \ at \ the \ centroid \ of \ triangle = 3.75 \times 10^7 \ V}}$

Given:

Charge (q) = +250 mC = $\rm 250 \times 10^{-6}$ C

Distance between centroid and corner of equilateral triangle (r) = 18 cm = 0.18 m

Explanation:

Electric potential due to a point charge (q) at a distance (r) is given as:

$\boxed{ \bold{V = \dfrac{kq}{r} }}$

$\rm k \longrightarrow \dfrac{1}{4 \pi \epsilon_o} = 9 \times 10^9$

So, resultant electric potential at centroid of the triangle due to charges at corners of triangle;

$\rm \implies V = \dfrac{kq}{r} + \dfrac{kq}{r} + \dfrac{kq}{r} \\ \\ \rm \implies V = 3\dfrac{kq}{r} \\ \\ \rm \implies V = \dfrac{3 \times 9 \times {10}^{9} \times 250 \times {10}^{ - 6} }{0.18} \\ \\ \rm \implies V = \frac{6750 \times {10}^{9 - 6} }{0.18} \\ \\ \rm \implies V = 37500 \times {10}^{3} \\ \\ \rm \implies V = 3.75 \times {10}^{7} \: V$

Answer 2

Answer:

According to the question,

We have an equilateral triangle. At each of its corner there is a charge of +250 mC placed in the air. We are to find the electric potential at the centroid of the equilateral triangle.

We can derive it using the given formula;

net potential at a centroid = ×

Ans) The resultant electric potential at the centroid =​ 3.75 ×