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Electrostatic Force


  • Electric forces between charged objects:
DIRECTION OF FORCE: Opposite charges attract each other and same charges repel each other!


MAGNITUDE OF FORCE: from Coulomb's Law

Coulomb’s law

  • The force between to charged particles is proportional to the product of the two charges, and inversely proportional to the distance squared.
Fe=keq1q2r2F_e=k_e\frac{|q_1||q_2|}{r^2}
where q1 and q2 correspond to the two charges, and ke is a proportionality constant equal to 8.99 x 109 N*m2*C-2.

  • Note that this equation only gives the magnitude of the force, hence the absolute values on the charges (so that the result is positive).
  • The direction of the force is determined by drawing a line joining both charges, and then deciding whether the force is attractive or repulsive. Then, the force is drawn accordingly along that line.

  • The superposition principle states that if there are more than two charged objects, the total force on one of these is obtained by calculating each individual force on that object (by taking the charges two by two) and then adding all the forces together.








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Electric Field

A charged particle with charge q, creates an electric field in their surrounding space.


  • The electric field is a vector field: for each point in space, you can define a vector quantity corresponding to that field.
  • It will have a magnitude and a direction.

  • The magnitude of the electric field created by a single point charge q is given by this formula:
E=keqr2E=k_e\frac{|q|}{r^2}

  • As for the direction of the electric field, it is radial, but we use the following convention: it comes out of a positive charge, and goes into a negative charge






  • If we introduce a test charged particle with charge qo in the region with an electric field, an electrostatic force (Fe) will be exerted on the test particle due to the presence of the electric field.
Fe=keqqor2Fe=qE\begin{array}{c} \displaystyle F_e=k_e\frac{|q||q_o|}{r^2}\\[11pt] \stackrel{\rightarrow}{F_e}=q\vec E \end{array}

  • Electric force is radial, meaning that it acts along the line joining two charges.

  • The force on a positive test charge particle is in the same direction of the electric field and the force on a negative charge particle is opposite the electric field direction.

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Two point charges are located as shown below. Where should be a third charge q3=8μCq_3=8\mu C placed on the x-axis so that the net force on q1q_1 isF1=8NF_1=8N in the negative x- direction?
Solution:
Since FnetF_{net}on q1q_1, is toward negative x and q3q_3 is a positive charge , q3q_3 should be on the right of q1q_1!

 F13F12=Fnet\rightarrow\ \left|\vec{F}_{13}\right|-\left|\vec{F}_{12}\right|=\left|F_{net}\right|
=k q1q3r13 2k q1q2r12 2=8=\left|\frac{k\ q_1q_3}{r_{13}^{\ 2}}\right|-\left|\frac{k\ q_1q_2}{r_{12}^{\ 2}}\right|=8

 r132=k q1 q38+k q1q2r122=9×109×2×8×10128+9×109×2×6×1012(0.3 )2=0.0157\rightarrow\ r_{13}^2=\frac{\left|k\ q_1\ q_3\right|}{8+\left|\frac{k\ q_1q_2}{r_{12}^2}\right|}=\frac{9\times10^9\times2\times8\times10^{^{-12}}}{8+\frac{9\times10^9\times2\times6\times10^{-12}}{\left(0.3\ \right)^2}}=0.0157

 r13=0.0157=0.125 m\rightarrow\ r_{13}=\sqrt{0.0157}=0.125\ mon the right of q1q_1
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A charged ball is placed 1m away from a sphere with a +1nC charge. How far away from a sphere with a +4nC charge would it have to be placed to be subjected to the same force?
Extra Practice