Question

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has magnitude of ????2=3.80 nC and is located at (x=1.50 m,y=0.650 m) , calculate the x and y components, ????x and ????y , of the electric field, ????⃗ , in component form at the origin, (0,0) . The Coulomb force constant is 1/(4????????0)=8.99×109 N⋅m2/C2 .

Answer 1

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Step-by-step explanation:

Given:

• Charge on first charged particle,
  `q_1=-4.10\ nC=-4.10* 10^(-9)\ C.`
• Charge on the second charged particle,
  `q_2=3.80\ nC=3.80* 10^(-9)\ C.`

• Position of the first charge =
  `(x_1=0.00\ m,\ y_1=0.600\ m).`
• Position of the second charge =
  `(x_2=1.50\ m,\ y_2=0.650\ m).`

The electric field at a point due to a charge
`q` at a point
`r` distance away is given by

`\vec E = (kq)/(|\vec r|^2)\ \hat r.`

where,

• 
  `k` = Coulomb's constant, having value
  `\rm 8.99* 10^9\ Nm^2/C^2.`

• 
  `\vec r` = position vector of the point where the electric field is to be found with respect to the position of the charge
  `q`.

• 
  `\hat r` = unit vector along
  `\vec r`.

The electric field at the origin due to first charge is given by

`\vec E_1 = (kq_1)/(|\vec r_1|^2)\ \hat r_1.`

`\vec r_1` is the position vector of the origin with respect to the position of the first charge.

Assuming,
`\hat i,\ \hat j` are the units vectors along x and y axes respectively.

`\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=(\vec r_1)/(|\vec r_1|)=(0.6\ \hat j)/(0.6)=-\hat j.`

Using these values,

`\vec E_1 = ((8.99* 10^9)* (-4.10* 10^(-9)))/((0.6)^2)\ (-\hat j)=1.025* 10^2\ N/C\ \hat j.`

The electric field at the origin due to the second charge is given by

`\vec E_2 = (kq_2)/(|\vec r_2|^2)\ \hat r_2.`

`\vec r_2` is the position vector of the origin with respect to the position of the second charge.

`\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = √((-1.5)^2+(-0.65)^2)=1.635\ m.\\\hat r_2=(\vec r_2)/(|\vec r_2|)=(-1.5\hat i-0.65\hat j)/(1.634)=-0.918\ \hat i-0.398\hat j.`

Using these values,

`\vec E_2= ((8.99* 10^9)* (3.80* 10^(-9)))/((1.635)^2)(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.`

The net electric field at the origin due to both the charges is given by

`\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.`

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.