Question

Two electric charges 3μC , -4μC are placed at the two corners of an isosceles right angled triangle of side 1m. What is the direction and magnitude of electric field at A due to the two charges?

Answer 1

Explanation:

E1 = 9 × 10^9 ( 3 × 10^-6 / 1^2 )

= 27 × 10^3 N/C along BA

E2 = 9× 10^9 ( 4 × 10^-6 )

= 36 × 10^3 N/C along AC

The angle between E1 and E2 is ( 180 - 45 ) = 135

Resultant electric field at A = E1 + E2

= ( E1^2 + E2^2 + 2E1.E2cosα )^1/2

{ ( 27 × 10^3 )^2 + ( 36 × 10^3 )^2 + 2× 27×10^3×36×10^3×cos135 }^1/2

= 25.50 × 10^3 N/C

Answer 2

Answer:

The Resultant Electric field at Point A due to charges is 25502.6  N/C   .

Explanation:

Given as :

Two electric charges 3μC , -4μC are placed at the two corners of an isosceles right angled triangle of side 1 m .

Let A B C is an isosceles triangle

The measure of side AB = 1 m

The measure of side BC = 1 m

The charge at point B = Q = 3 μC

The charge at point C = q = - 4 μC

The Electric field direction through side BA = $\vec{E_1}$

The Electric field direction through side AC = $\vec{E_2}$

According to question

Electric field along BA

$\vec{E_1}$ = k $\dfrac{Q}{r^{2} }$

Or, $\vec{E_1}$ =  9 ×  $10^{9}$ × $\dfrac{3\times 10^{-6}}{1^{2}}$

∴    $\vec{E_1}$ = 27  ×  $10^{9-6}$

i.e   $\vec{E_1}$ = 27  ×  $10^{3}$  N/C  along BA

Electric field at point B =  27  ×  $10^{3}$  N/C  along BA

Similarly

Electric field along AC

$\vec{E_2}$ =  9 ×  $10^{9}$ × $\dfrac{4\times 10^{-6}}{1^{2}}$

∴    $\vec{E_2}$ = 36  ×  $10^{9-6}$

i.e   $\vec{E_2}$ = 36 ×  $10^{3}$  N/C  along AC

Electric field at point C =  $\vec{E_2}$ = 36 ×  $10^{3}$  N/C  along AC

Now, From figure it is clear that ,angle between electric filed vectors = ( 180° - 45° )

i.e

Angle between electric filed vectors $\vec{E_1}$  and $\vec{E_2}$  = α = 135°

Now,

Resultant Electric field at Point A due to charge  = $\vec{E_1}$  + $\vec{E_2}$

Or,  $\vec{E_1}$  + $\vec{E_2}$  = $\sqrt{E_1^{2}+E_2^{2}+2E_1E_2cos\alpha }$     (from Parallelogram Law of force )

i.e  $E_r$  = $\sqrt{(27000)^{2}+(36000)^{2}+2\times 27000\times 36000cos135^{\circ}}$

∴   $E_r$  = $\sqrt{650384417}$

i.e $E_r$  = 25502.6  N/C

So, The Resultant Electric field at Point A due to charge  = $E_r$  = 25502.6  N/C

Hence, The Resultant Electric field at Point A due to charges is 25502.6  N/C   . Answer