Question

Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 4.25×10^−6 C charges. Express your answer to three significant figures and include the appropriate units.

Answer 1

`1.250* 10^4\ N/C.`

Step-by-step explanation:

Given:

• Charge on each corner of the square,
  `q=4.25* 10^(-6)\ C.`
• Length of the side of the square,
  `a = 2.42\ m.`

According to the Coulomb's law, the strength of the electric field at a point due to a charge
`q` at a point
`r` distance away is given by

`E = (kq)/(r^2)`

where,

`k` = Coulomb's constant =
`9* 10^9\ Nm^2/C^2`.

The direction of the electric field is along the line joining the point an d the charge.

The electric field at the point P due to charge at A is given by

`E_A = (kq)/(a^2)`

Since, this electric field is along positive x axis direction, therefore,

`\vec E_A = (kq)/(a^2)\ \hat i.`

`\hat i` is the unit vector along the positive x-axis direction.

The electric field at the point P due to charge at B is given by

`E_B = (kq)/(a^2)`

Since, this electric field is along negative y axis direction, therefore,

`\vec E_B = (kq)/(a^2)\ (-\hat j).`

`\hat j` is the unit vector along the positive y-axis direction.

The electric field at the point P due to charge at C is given by

`E_C = (kq)/(r^2)`

where,
`r=√(a^2+a^2)=a\sqrt 2`.

Since, this electric field is along the direction, which is making an angle of
`45^\circ` below the positive x-axis direction, therefore, the direction of this electric field is given by
`\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)`.

`\vec E_C = (kq)/(2a^2)\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=(kq)/(2a^2)\ ((1)/(\sqrt 2)\hat i-(1)/(\sqrt 2)\hat j)\\=(kq)/(2a^2)\ (1)/(\sqrt 2)(\hat i-\hat j).\\`

Thus, the total electric field at the point P is given by

`\vec E = \vec E_A+\vec E_B +\vec E_C\\=(kq)/(a^2)\hat i+(kq)/(a^2)(-\hat j)+(kq)/(2a^2)\ (1)/(\sqrt 2)(\hat i-\hat j).\\=\left ( (kq)/(a^2)+(kq)/(2\sqrt 2\ a^2) \right )\hat i+\left ((kq)/(a^2)+(kq)/(2\sqrt 2\ a^2) \right )(-\hat j)\\=(kq)/(a^2)\left [\left ( 1+(1)/(2\sqrt 2) \right )\hat i+\left (1+(1)/(2\sqrt 2)  \right )(-\hat j)\right ]\\=(kq)/(a^2)(1.353\hat i-1.353\hat j)`

`=((9* 10^9)* (4.25* 10^(-6)))/(2.42^2)* (1.353\hat i-1.353\hat j)\\=(8.837* 10^3\hat i-8.837* 10^3\hat j)\ N/C.`

The magnitude of the electric field at the given point due to all the three charges is given by

`E=√((8.837* 10^3)^2+(-8.837* 10^3)^2)=1.250* 10^4\ N/C.`

Answer 2

Ans:

12500 N/C

Step-by-step explanation:

Side of square, a = 2.42 m

q = 4.25 x 10^-6 C

The formula for the electric field is given by

`E = (Kq)/(r^2)`

where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges

According to the diagram

BD =
`√(2)* a`

where, a be the side of the square

So, Electric field at B due to charge at A

`E_(A)=(Kq)/(a^2)`

`E_(A)=(9*10^(9)* 4.25 * 10^(-6))/(2.42^2)`

EA = 6531.32 N/C

Electric field at B due to charge at C

`E_(C)=(Kq)/(a^2)`

`E_(C)=(9*10^(9)* 4.25 * 10^(-6))/(2.42^2)`

Ec = 6531.32 N/C

Electric field at B due to charge at D

`E_(D)=(Kq)/(2a^2)`

`E_(D)=(9*10^(9)* 4.25 * 10^(-6))/(2* 2.42^2)`

ED = 3265.66 N/C

Now resolve the components along X axis and Y axis

Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C

The resultant electric field at B is given by

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

`E=\sqrt{8840.5^(2)+8840.5^(2)}`

E = 12500 N/C

Step-by-step explanation: