Physics Help, Electric Field Exercises. Point and Distributed Charges.

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Exercise one, Physics Help. Electric Field Created by Point Charges.

Find the electric field at point P on the y axis and at a distance 0.4 m above the origin. The electric field is created by the three point charges as shown below. The charge q₁ = 7x10^-6 C is in the origin of the coordinate system, the charge q₂ = -5x10^-6 C is placed on the x axis 0.3 m from the origin and the charge q₃ = -3x10^-6 C is placed to the right of point P and at 0.4 m above q₂.
Determine also the force exerted on a charge of 3x10^-6 C when placed on the point P.

SOLUTION:
Let's first calculate the magnitude of the electric field at P due to the presence of each charge. Denote E₁ the electric field produced by q₁, E₂ the electric field produced by q₂ and E₃ the electric field produced by q₃. These electric fields are represented in the figure and their magnitudes are, using positive values for the charges:

E₁ = keq₁/r₁² = (9x10⁹ Nm²/C²)(7.0x10^-6 C)/(0.4m)²
= 3.9x10⁵ N/C.
E₂ = keq₂/r₂² = (9x10⁹ Nm²/C²)(5.0x10^-6 C)/(0.5m)²
= 1.8x10⁵ N/C.
E₃ = keq₃/r₃² = (9x10⁹ Nm²/C²)(3.0x10^-6 C)/(0.3m)²
= 3.0x10⁵ N/C.

Vector E₁ has no x component, just y component (upward). Vector E₂ has a x component given by E₂cosß = (3/5)E₂ and a negative y component given by -E₂ senß = -(4/5)E₂. Vector E₃has no y component, just x component (to the right).

The resultant vector E we are looking for is the vectorial sum of this three vectors, E = E₁ + E₂ + E₃. The vectors E₁, E₂ and E₃ are denoted using unitary vectors i and j. Then we can obtain their sum analytically.

E₁ = 3.9x10⁵ j N/C.
E₂ = (1.1x10⁵ i - 1.4x10⁵ j )N/C.
E₃ = 3.0x10⁵ i N/C.

The electric field E at P is then:

E = (4.1x10⁵ i + 2.5x10⁵ j) N/C.

The force on a charge of 3x10^-6 C when is placed at the point P is obtained using F = Eq, with q = 3x10^-6 C, F = (1.23 i + 0.75 j)N. This force has of course the same direction of the electric field E, because the charge q is positive.

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Physics Help, Exercise Two, Electric Field. Distributed charges.

In the figure a thin rod of length L and a total charge Q uniformly distributed is placed along the x axis. Find the magnitude of the electric field at point A located a distance d from the left side of the rod.

This problem can be solved assuming the rod is made of point charges of magnitude dq = (Q/L)dx. Q/L is the charge per unit length. We can use E = Kq/x² in differential form, dE = Kdq/x². dE is the contribution to the field of dq at a distance x from A. Then we add the contributions of all the electric point charges using limits of integration from d to d + L: