Question

Three charges of respective values - sqrt(2) mu C 2sqrt(2)mu C and -sqrt(2)mu C are arranged along a straight line as shown in the figure.Calculate the total electric field intensity due to all three charges at the point P .​

Answer 1

Given that,

First charge $q_{1}=\sqrt{2}\ \mu C$

Second charge $q_{2}=2\sqrt{2}\ \mu C$

Third charge $q_{3}=\sqrt{2}\ \mu C$

According to figure,

We need to calculate the electric field at point P due to charge at A

Using formula of electric field

$E_{PA}=\dfrac{kq_{1}}{r^2}$

Where, q₁ = first charge

r = distance

Put the value into the formula

$E_{PA}=\dfrac{9\times10^{9}\times\sqrt{2}\times10^{-6}}{(\sqrt{2})^2}$

$E_{PA}=\dfrac{9\sqrt{2}}{2}\times10^{3}\ N/C$

We need to calculate the electric field at point P due to charge at C

Using formula of electric field

$E_{PC}=\dfrac{kq_{3}}{r^2}$

Where, q₃ = third charge

r = distance

Put the value into the formula

$E_{PC}=\dfrac{9\times10^{9}\times\sqrt{2}\times10^{-6}}{(\sqrt{2})^2}$

$E_{PC}=\dfrac{9\sqrt{2}}{2}\times10^{3}\ N/C$

We need to calculate the electric field at point P due to charge at B

Using formula of electric field

$E_{PB}=\dfrac{kq_{2}}{r^2}$

Where, q₂ = first charge

r = distance

Put the value into the formula

$E_{PB}=\dfrac{9\times10^{9}\times2\sqrt{2}\times10^{-6}}{(1)^2}$

$E_{PB}=18\sqrt{2}\times10^{3}\ N/C$

We need to calculate the horizontal component of net electric field at point P

Using horizontal component

$E_{x}=E_{PC}\cos\theta-E_{PA}\cos\theta$

$E_{x}=0$

We need to calculate the vertical component of net electric field at point P

Using vertical component

$E_{y}=E_{PB}-E_{PA}\sin\theta-E_{AC}\sin\theta$

$E_{y}=E_{PB}-E_{PA}\sin 45-E_{AC}\sin 45$

Put the value into the formula

$E_{y}=(18\sqrt{2}-\dfrac{9\sqrt{2}}{2}\times\dfrac{1}{\sqrt{2}}-\dfrac{9\sqrt{2}}{2}\times\dfrac{1}{\sqrt{2}})\times10^{3}$

$E_{y}=16.45\times10^{3}\ N/C$

We need to calculate the total electric field at point P

Using formula of electric field

$E=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E=\sqrt{0+(16.45\times10^{3})^2}$

$E=16450\ N/C$

Hence, The total electric field at point P is 16450 N/C.