The figure shows an uneven arrangement of electrons e and pr

The figure shows an uneven arrangement of electrons (e) and protons (p) on a circular arc of radius r = 1.79 cm, with angles ₁ = 24°, ₂ = 50°, ₃ = 30°,₄ = 20°. What are the magnitude and direction (relative to the positive direction of the x axis) of the net electric field produced at the center of the arc?

(a) Magnitude
N/C

(b) Direction
°

Solution

For a charge Q, the electric field at a distance r is given as E = kq/r²

Also, for a positive charge, the direction of the field is away from the charge while for a negative charge the field directs towards the charge.

Now for the given problem, we will make use of the above and then find the electric field along the x and y directions separately and then add up to find the net magnitude.

Hence, for X direction:

Ex = (K/r²)[e - pCos24 + pCos74 - eCos50 + pCos20]

or, Ex = (k/r²) (1.60217662 × 10^-19)[1 - Cos24 + Cos74 - Cos50 + Cos20]

or, Ex = 8.99 x 10^9(1.60217662 × 10^-19)(1+0.27564 + 0.93969 - 0.91355 - 0.64279) x 100 / 1.79*1.79

or, Ex = 296.239 x 10^-10 N/C

Again, for the Y direction we have:

Ey = 8.99 x 10^9(1.60217662 × 10^-19)x100( -Sin24 - Sin74 + Sin50 - Sin20) / 1.79*1.79

or, Ey = 449.535 x 10^-10x ( -Sin24 - Sin74 + Sin50 - Sin20)

or, Ey = 449.535 x 10^-10 x (0.766 - 0.40674 - 0.96126 - 0.342)

or, Ey = 449.535 x 10^-10 (-0.944) = -424.36104 x 10^-10 N/C

Therefore the net magnitude of the electric would be given as: E = sqrt(Ex² + Ey²)

Part a.) That is E = sqrt[296.239² + 424.36104²] x 10^-10 =517.5325 x 10^-10 N/C

Part b.) For the direction of the net electric field, we can write:

Tan = -424.36104 / 296.239

or, = -55.0818

Therefore the net electric field makes an angle of 55.0818 with +x axis in the clockwise direction.