electric potential in a region of space is given by v =x^2+ y^2+ z^2.what will be the electric field intensity at a point p (- 1 m ,2m, -3 m)
Answers
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Explanation:
The electrostatic force is a conservative force. This means that the work it does on a particle depends only on the initial and final position of the particle, and not on the path followed. With each conservative force, a potential energy can be associated. The introduction of the potential energy is useful since it allows us to apply conservation of mechanical energy which simplifies the solution of a large number of problems.
The potential energy U associated with a conservative force F is defined in the following manner
(25.1)
where U(P0) is the potential energy at the reference position P0 (usually U(P0) = 0) and the path integral is along any convenient path connecting P0 and P1. Since the force F is conservative, the integral in eq.(25.1) will not depend on the path chosen. If the work W is positive (force and displacement pointing in the same direction) the potential energy at P1 will be smaller than the potential energy at P0. If energy is conserved, a decrease in the potential energy will result in an increase of the kinetic energy. If the work W is negative (force and displacement pointing in opposite directions) the potential energy at P1 will be larger than the potential energy at P0. If energy is conserved, an increase in the potential energy will result in an decrease of the kinetic energy. If In electrostatic problems the reference point P0 is usually chosen to correspond to an infinite distance, and the potential energy at this reference point is taken to be equal to zero. Equation (25.1) can then be rewritten as:
To describe the potential energy associated with a charge distribution the concept of the electrostatic potential V is introduced. The electrostatic potential V at a given position is defined as the potential energy of a test particle divided by the charge q of this object:
(25.3)
In the last step of eq.(25.3) we have assumed that the reference point P0 is taken at infinity, and that the electrostatic potential at that point is equal to 0. Since the force per unit charge is the electric field (see Chapter 23), eq. (25.3) can be rewritten as
(25.4)
The unit of electrostatic potential is the volt (V), and 1 V = 1 J/C = 1 Nm/C. Equation (25.4) shows that as the unit of the electric field we can also use V/m.
A common used unit for the energy of a particle is the electron-volt (eV) which is defined as the change in kinetic energy of an electron that travels over a potential difference of 1 V. The electron-volt can be related to the Joule via eq.(25.3). Equation (25.3) shows that the change in energy of an electron when it crosses over a 1 V potential difference is equal to 1.6 . 10-19 J and we thus conclude that 1 eV = 1.6 . 10-19 J
25.2. Calculating the Electrostatic Potential
A charge q is moved from P0 to P1 in the vicinity of charge q' (see Figure 25.1). The electrostatic potential at P1 can be determined using eq. (25.4) and evaluating the integral along the path shown in Figure 25.1. Along the circular part of the path the electric field and the displacement are perpendicular, and the change in the electrostatic potential will be zero. Equation (25.4) can therefore be rewritten as
(25.5)
If the charge q' is positive, the potential increases with a decreasing distance r. The electric field points away from a positive charge, and we conclude that the electric field points from regions with a high electrostatic potential towards regions with a low electrostatic potential.
Figure 25.1. Path followed by charge q between P0 and P1.
From the definition of the electrostatic potential in terms of the potential energy (eq.(25.3)) it is clear that the potential energy of a charge q under the influence of the electric field generated by charge q' is given by
".
Figure 25.4. Cross section of cavity inside spherical conductor.
Figure 25.4 shows the cross section of a possible cavity inside a spherical conductor. Suppose there is a field inside the conductor and one of the field lines is shown in Figure 25.4. Consider the path integral of eq.(25.16) along the path indicated in Figure 25.4. In Chapter 24 it was shown that the electric field within a conductor is zero. Thus the contribution of the path inside the conductor to the path integral is zero. Since the remaining part of the path is chosen along the field line, the direction of the field is parallel to the direction of the path, and therefore the path integral will be non-zero. This obviously violates eq.(25.16) and we must conclude that the field inside the cavity is equal to zero (in this case the path integral is of course equal to zero).
25.4. The Gradient of the Electrostatic Potential
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