Question

two point charges + q and minus 2 q are placed at the vertices B and C of an equilateral triangle of side is obtained and expression for the magnitude and direction of the resultant electron field at a due to these two charges​

Answer 1

Answer:

b c are two difht of class 12

Answer 2

Given:

Two point charges + q and minus 2 q are placed at the vertices B and C of an equilateral triangle.

To find:

Resultant field intensity at point A ?

Calculation:

• At point A , angle between field intensity vectors (from charges at point B and C) are at angle 60°.

• Field intensity due to q charge = kq/a².

• Field intensity due to -2q charge = k(2q)/a².

Now, these vectors are at 60° angle.

$\sf \: E = \sqrt{ { \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2} + { \bigg( \dfrac{2kq}{ {a}^{2} } \bigg)}^{2} + 2.\bigg( \dfrac{kq}{ {a}^{2} } \bigg).\bigg( \dfrac{2kq}{ {a}^{2} } \bigg) \cos( {60}^{ \circ} ) }$

$\sf \implies \: E = \sqrt{ { \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2} + { \bigg( \dfrac{2kq}{ {a}^{2} } \bigg)}^{2} + \bigg( \dfrac{kq}{ {a}^{2} } \bigg).\bigg( \dfrac{2kq}{ {a}^{2} } \bigg)}$

$\sf \implies \: E = \sqrt{ { \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2} + 4 { \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2} + 2{ \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2}}$

$\sf \implies \: E = \sqrt{ 7{ \bigg( \dfrac{kq}{ {a}^{2} } \bigg)}^{2} }$

$\sf \implies \: E = \dfrac{kq\sqrt{ 7 }}{ {a}^{2} }$

Hope It Helps.