**Electric
Field, Physics Theory: Press Here**

**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 q _{1} = 7x10^{-6}
C is in the origin of the coordinate system, the charge q_{2} = -5x10^{-6}
C is placed on the x axis 0.3 m from the origin and the charge q_{3}
= -3x10^{-6} C is placed to the right of point P and at 0.4 m above
q_{2}**

Determine also the force exerted on a charge of 3x10

**SOLUTION:
Let's first calculate the magnitude of the electric field at P due to the
presence of each charge. Denote E _{1} the electric field produced
by q_{1}, E_{2} the electric field produced by q_{2}
and E_{3} the electric field produced by q_{3}. These electric
fields are represented in the figure and their magnitudes are, using positive
values for the charges:
E_{1} = keq_{1}/r_{1}^{2} = (9x10^{9}
Nm^{2}/C^{2})(7.0x10^{-6} C)/(0.4m)^{2}
= 3.9x10^{5}
N/C.
E_{2} = keq_{2}/r_{2}^{2} = (9x10^{9}
Nm^{2}/C^{2})(5.0x10^{-6} C)/(0.5m)^{2}
= 1.8x10^{5}
N/C.
E_{3} = keq_{3}/r_{3}^{2} = (9x10^{9}
Nm^{2}/C^{2})(3.0x10^{-6} C)/(0.3m)^{2}
= 3.0x10^{5}
N/C. **

**Vector E _{1} has no x component,
just y component (upward). Vector E_{2} has a x component given by
E_{2}cosß = (3/5)E_{2} and a negative y component given by
-E_{2} senß = -(4/5)E_{2}. Vector E_{3 }has 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 = E _{1} + E_{2} + E_{3}. The vectors E_{1},
E_{2} and E_{3} are denoted using unitary vectors i and j.
Then we can obtain their sum analytically.**

E

E

**The electric field E at P is then:**

**E = (4.1x10 ^{5} i + 2.5x10^{5} 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. **

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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/x ^{2} in differential form, dE
= Kdq/x^{2}. 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: **

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