Question

Three charges, q₁ = +2.60nC,q₂ = -1.48nC, and q₃ = +8.89nC, are at the corners of an equilateral triangle, each of whose sides are of length, L, as shown in the figure below.

The angle α is 60.0° and L = 0.407 m. We are interested in the unmarked point midway between the chargesq₁ and q₂ on the xaxis.

For starters, calculate the magnitude and direction of the electric field due only to charge q₁ at this point.

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Incorrect.

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Calculate the magnitude and direction of the electric fielddue only to charge q₂ at this point.

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Calculate the magnitude and direction of the electric fielddue only to charge q₃ at this point.

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Now calculate the magnitude of the electric field from all three charges at a point midway between the two charges on thex axis.

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Calculate the angle of the electric field relative to the positive (to the right) x-axis, with positive values up (counterclock-wise) and negative down (clockwise). (enter the answer with units of deg)

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If a tiny particle with a charge q= 1.48nC were placed at this point midway between q₁ andq₂, what is the magnitude of the force it would feel?

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This discussion is closed.

Answer 1

Charge is located at ( at origin)

Charge is located at

Charge is located at

Electric field is to be determined at point $vec{r}=rac{L}{2}hat{i}$ (midway between charges 1 and 2)

Electric field at the required point due to the charge 1 is

$vec{E}_1=rac{q_1(vec{r}-vec{r}_1)}{4piepsilon_0|vec{r}-vec{r}_1|^3}$

$vec{E}_1=564.4hat{i},N/C$

Magnitude of electric field due to charge 1 is and its direction is towards positive X-axis.(right)

El = 564.4 yc

Electric field at the required point due to the charge 2 is

$vec{E}_2=rac{q_2(vec{r}-vec{r}_2)}{4piepsilon_0|vec{r}-vec{r}_2|^3}$

$vec{E}_2=321.3hat{i},N/C$

Magnitude of electric field due to charge 2 is ${E}_2=321.3,N/C$ and its direction is towards positive X-axis.(right)

Electric field at the required point due to the charge 3 is

$vec{E}_3=rac{q_3(vec{r}-vec{r}_3)}{4piepsilon_0|vec{r}-vec{r}_3|^3}$

$vec{E}_3=-643.3hat{j},N/C$​​​​​​​

Magnitude of electric field due to charge 3 is ${E}_3=643.3,N/C$ ​​​​​​​ and its direction is towards negative Y-axis.(down)

Net electric field is $vec{E}=(885.7hat{i}-643.3hat{j}),N/C$

Magnitude of net electric field is ${E}=sqrt{885.7^2+643.3^2}=1094.7,N/C$

Direction of net electric field is counter-clockwise with positive X-axis.

If a particle with charge $q=1.48,nC$ is placed at the required point,

Force on the particle is $F=qE=1.48*10^{-9}*1094.7=1.62*10^{-6},N$