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18.6 THE ELECTRIC FIELD There may be many charges that give an object an electrostatic force. Picture on 547 Call qo a test charge: for determining the extent to which the surrounding charges generate a force.
The positive test charge shown in Figure 18.15 is qo = +3.0 x 10^-8 C and experiences a force F = 6.0 x 10^-8 N in the direction shown in the drawing.
The charges in the environment apply a force F to the test charge qo. The force per coulomb experienced by the test charge is F/qo. If qo is replaced by a new charge q, then the force on this new charge is the force per coulomb times q.
(a) The force per coulomb of charge is
(b) The result from part (a) indicates that the surrounding charges can exert 2.0 newtons of force per coulomb of charge. Thus, a charge of +12 x 10^-8 C would experience a force whose magnitude is
The direction of this force would be the same as that experienced by the test charge, since both have the same positive sign.
The electric force per coulomb F/qo, calculated in Ex. 6(a) illustrates the electricfield.
Definition of the Electric Field
The electric field E that exists at a point is the electrostatic force F experienced by a small test charge qo placed at the point divided by the charge itself:
The electric field is a vector, and its direction is the same as the direction of the force F on a positive test charge.
It is the surrounding charges that create an electric field at a given point.
fig. 18.16a Charges on two metal spheres and the ebonite rod create an electric field E at the spot indicated. This field has a magnitude of 2.0 N/C and is directed as in the drawing. Determine the force on a charge placed at that spot, if the charge has a value of (a) qo = +18 x 10^-8C and (b) qo = -24 x 10^-8C.
The electric field at a given spot can exert a variety of forces, depending on the magnitude and sign of the charge placed there. The charge is assumed to be small enough that it does not alter the locations of the surrounding charges that create the field.
Since qo is positive, the force points in the same direction as the electric field, as part a of the drawing indicates.
The force on the negative charge points in the direction opposite to the force on the positive charge – that is, opposite to the electric field (part b drawing).
Each surrounding charge contributes to the net electric field that exists.
When two or more waves are present simultaneously at the same place, the resultant disturbance is the sum of the disturbances from the individual waves.
Fig. 18.17 two charges objects, A and B. Each contributes as follows to the net electric field at point P: Ea = 3.00N/C directed to the right, and Eb = 2.00N/C directed downward. Thus Ea and Eb are perpendicular. What is the net electric field at P?
The net electric field E is the vector sum of Ea and Eb: E = Ea + Eb. Ea and Eb are perpendicular, so E is the diagonal of the rectangle shown in the drawing. Use the Pythagorean theorem to find the mag. Of E and trig to find the directional angle.
In a vacuum, a proton (m = 1.67 x 10^-27kg) is moving parallel to a uniform electric field that is directed along the +x axis. The proton starts with a velocity of +2.5 x 10^4m/s and accelerates in the same direction as the electric field, which has a value of +2.3 x 10^3N/C. Find the velocity of the proton when its displacement is +2.0 mm from the starting point.
There is an isolated point charge of q=+15 uC in a vacuum. Using a test charge of qo = +.80 uC, determine the electric field at point P, which is 0.20 m away.
Electric field produced by a point charge can be obtained in general terms from Coulomb’s law.