Electric field strength / intensity at a point is defined as the force
per unit positive charge acting at that point {a vector; Unit: N C
^{1} or V m
^{1}}
E = F / q → F = qE
 The electric force on a positive charge in an electric field is in the direction of E, while
 The electric force on a negative charge is opposite to the direction of E.
 Hence a +ve charge placed in an electric field will accelerate in the direction of E and gain KE {& simultaneously lose EPE}, while a negative charge caused to move (projected) in the direction of E will decelerate, ie lose KE, { & gain EPE}.
Representation of electric fields by field lines
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[/t] Coulomb's law: The (mutual) electric force F acting between 2 point charges Q_{1} and Q_{2} separated by a distance r is given by:
F = Q_{1}Q_{2} / 4πε_{o}r^{2} where ε_{0}: permittivity of free space or, the (mutual) electric force between two point charges is proportional to the product of their charges & inversely proportional to the square of their separation. Example 1: Two positive charges, each 4.18 μC, and a negative charge, 6.36 μC, are fixed at the vertices of an equilateral triangle of side 13.0 cm. Find the electrostatic force on the negative charge.  F = Q_{1}Q_{2} / 4πε_{o}r^{2} = (8.99 x 1 0^{9}) [(4.18 x 10^{6})(6.36 x 10^{6}) / (13.0 x 10^{2})^{2}] = 14.1 N (Note: negative sign for 6.36 μC has been ignored in the calculation) F_{R} = 2 x Fcos30^{0} = 24.4 N, vertically upwards  Electric field strength due to a Point Charge Q : E = Q / 4πε_{o}r^{2} {NB: Do NOT substitute a negative Q with its negative sign in calculations!} Example 2: In the figure below, determine the point (other than at infinity) at which the total electric field strength is zero. From the diagram, it can be observed that the point where E is zero lies on a straight line where the charges lie, to the left of the 2.5 μC charge. Let this point be a distance r from the left charge. Since the total electric field strength is zero, E_{6μ} = E_{2μ} [6μ / (1 + r)^{2}] / 4πε_{o}r^{2} = [2.5μ / r^{2}] / 4πε_{o}r^{2} (Note: negative sign for 2.5 μC has been ignored here) 6 / (1 + r)^{2} = 2.5 / r^{2} √(6r) = 2.5 (1 + r) r = 1.82 m The point lies on a straight line where the charges lie, 1.82 m to the left of the 2.5 μC charge.
Uniform electric field between 2 Charged Parallel Plates: E = Vd, d: perpendicular dist between the plates, V: potential difference between plates Path of charge moving at 90° to electric field: parabolic. Beyond the pt where it exits the field, the path is a straight line, at a tangent to the parabola at exit. Example 3: An electron (m = 9.11 x 10^{31} kg; q = 1.6 x 10^{19} C) moving with a speed of 1.5 x 10^{7} ms^{1}, enters a region between 2 parallel plates, which are 20 mm apart and 60 mm long. The top plate is at a potential of 80 V relative to the lower plate. Determine the angle through which the electron has been deflected as a result of passing through the plates. Time taken for the electron to travel 60 mm horizontally = Distance / Speed = 60 x 1 0^{3} / 1.5 x 10^{7} = 4 x 10^{9} s E = V / d = 80 / 20 x 10^{3} = 4000 V m^{1} a = F / m = eE / m = (1.6 x 10^{19})(4000) / (9.1 x 10^{31}) = 7.0 x 10^{14} ms^{2} v_{y} = u_{y} + at = 0 + (7.0 x 10^{14})( 4 x 10^{9}) = 2.8 x 10^{6} ms^{1} tan θ = v_{y} / v_{x} = 2.8 x 10^{6} / 1.5 x 10^{7} = 0.187 Therefore θ = 10.6° Effect of a uniform electric field on the motion of charged particles  Equipotential surface: a surface where the electric potential is constant
 Potential gradient = 0, ie E along surface = 0 }
 Hence no work is done when a charge is moved along this surface.{ W=QV, V=0 }
 Electric field lines must meet this surface at right angles.
 {If the field lines are not at 90° to it, it would imply that there is a nonzero component of E along the surface. This would contradict the fact that E along an equipotential = 0. }
Electric potential at a point: is defined as the work done in moving a unit positive charge from infinity to that point, { a scalar; unit: V } ie V = W / Q The electric potential at infinity is defined as zero. At any other point, it may be positive or negative depending on the sign of Q that sets up the field. {Contrast gravitational potential.} Relation between E and V: E =  dV / dr i.e. The electric field strength at a pt is numerically equal to the potential gradient at that pt. NB: Electric field lines point in direction of decreasing potential {ie from high to low pot}. Electric potential energy U of a charge Q at a pt where the potential is V: U = QV Work done W on a charge Q in moving it across a pd ΔV: W = Q ΔV Electric Potential due to a point charge Q : V = Q / 4πε_{o}r {NB: Substitute Q with its sign} 

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