Question

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the square what would be the magnitude and direction of the net electric field a one of the other corners of the square?

Answer 1

The magnitude of the electric field and direction of electric field are
`146.03*10^(3)\ N/C` and 75.36°.

Step-by-step explanation:

Given that,

First charge
`q_(1)= 2.9\mu C`

Second charge
`q_(2)= 2.9\mu C`

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

`E_(1)=(kq)/(r'^2)`

Put the value into the formula

`E_(1)=(9*10^(9)*2.9*10^(-6))/((50√(2)*10^(-2))^2)`

`E_(1)=52200=52.2*10^(3)\ N/C`

For E₂,

Using formula of electric field

`E_(1)=(kq)/(r^2)`

Put the value into the formula

`E_(2)=(9*10^(9)*2.9*10^(-6))/((50*10^(-2))^2)`

`E_(2)=104400=104.4*10^(3)\ N/C`

We need to calculate the horizontal electric field

`E_(x)=E_(1)\cos\theta`

`E_(x)=52.2*10^(3)*\cos45`

`E_(x)=36910.97=36.9*10^(3)\ N/C`

We need to calculate the vertical electric field

`E_(y)=E_(2)+E_(1)\sin\theta`

`E_(y)=104.4*10^(3)+52.2*10^(3)\sin45`

`E_(y)=141310.97=141.3*10^(3)\ N/C`

We need to calculate the net electric field

`E_(net)=\sqrt{E_(x)^2+E_(y)^2}`

Put the value into the formula

`E_(net)=\sqrt{(36.9*10^(3))^2+(141.3*10^(3))^2}`

`E_(net)=146038.69\ N/C`

`E_(net)=146.03*10^(3)\ N/C`

We need to calculate the direction of electric field

Using formula of direction

`\tan\theta=(141.3*10^(3))/(36.9*10^(3))`

`\theta=\tan^(-1)((141.3*10^(3))/(36.9*10^(3)))`

`\theta=75.36^(\circ)`

Hence, The magnitude of the electric field and direction of electric field are
`146.03*10^(3)\ N/C` and 75.36°.