Set 51 Problem number 17


Problem

Problem: We refer to the number of volts / meter (or alternatively the number of Newtons/Coulomb) at a point as the electric field strength at that point.   In a certain region, the electric field is uniformly equal to 34 volts / meter.

Find  

the force exerted on a charge of 36 `microC in the vicinity of this field 

the work done by the field in moving this charge .1800 meters in the direction of the field 

the potential gradient dV/dx of the field 

the force per unit charge exerted by the field." 

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Solution

Solution:  An electric field of 34 volts/meter is equivalent to 34 N / C. Thus the 36 `microC charge will experience a force of  

36 `microC ( 34 N/C) = .001224 N. 

The work done over a .1800 meter displacement in the direction of the field is   

( .001224 N)( .1800 meters) = .0002203 Joules. 

The potential gradient is just the 34 volts / meter; the potential gradient and the electric field strength are identical. 

The force per unit charge is just 34 N/C. The electric field strength is the force per unit charge.

Generalized Solution

Generalized Response: A charge q in an electric field E will experience a force of q E, as is obvious from the units (E is measured in volts/meter, the same as Newtons/Coulomb, while q is measured in Coulombs). The work done in moving the charge through a displacement ds in the direction of the field will have magnitude |W| = |q E | ds. The potential difference has magnitude |W| / q = | E | ds. The potential gradient is potential difference/displacement, so has magnitude | E | ds / ds = | E |. Since E measures Newtons/Coulomb, it measures force/unit charge.

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Explanation in terms of Figure(s), Extension

Figure description: The figure below shows a charge q moving through a displacement `ds in the direction of a potential gradient dV / dx. The potential difference will be the product of the potential gradient and the displacement: `dV = dV / dx * `ds. The work done on the charge is therefore `dW = q `dV; this work is also the product `dW = Fave * `ds of the average force on the charge and the displacement. Setting these two expressions for work `dW equal we arrive at the conclusion that Fave = q dV/dx = q E.

E = dV / dx is called the electric field. Since E = Fave / q, we see that the electric field E is the force per unit charge experienced by a charge.

Figure(s)

potential_gradient_and_work.gif (3886 bytes)