Strategy
- Calculate the work done by an external force to move a test charge against the
constant force of the field. - State the work done on the field in a different way, using the relationship of
potential difference, work done on the field, and charge. - Set the two expressions for work equal to each other and simplify to get the
desired equation.
Physics principles and equations
The work done to move a test charge from point P 1 to point P 2 , opposite to the
direction of ǻs, is F·(íǻs). Since F is constant and parallel to the displacement
between the points, this dot product can be written as the scalar product Fǻs cos 0°,
or just
W = Fǻs
The force exerted on the test charge by the electric field is
F = qE
A relationship stated and derived earlier for the relationship between potential difference
and work done on a system is
ǻV = W/q
Step-by-step derivation
We find two expressions for the work W and set them equal to each other. We derive
this using a positive charge being pushed against the field. The work done by the
external force is positive in this case; work done by the field is negative.
Although we derived Equation 1 for the case of a positive charge being moved against a uniform field, the equation is equally true for negative
charges, and for charges moving in either direction.
What are the magnitude and
direction of this uniform electric
field?
Field is directed to the right
ǻs directed from left to right
ǻV = 10 V í 15 V = í 5.0 V
E = í (í 5.0 V)/(0.20 m) = 25 N/C
E = 25 N/C to the right
Step Reason
1. W = Fǻs definition of work
2. Ffield = qtestE electric field force on test charge
3. Fext = – qtestE external force on test charge
4. Wext = – qtestEǻs substitute equation 3 into equation 1
5. Wext = qtestǻV potential difference and work
6. ǻV = – Eǻs equate steps 4 and 5 and simplify
24.12 - Interactive checkpoint: an accelerated proton
In a uniform electric field, the potential
changes from 14.0 V to 32.6 V in a
displacement of 0.130 m (the
direction is opposite to the field, as
shown). What is the magnitude of the
acceleration of a proton (charge
1.60×10í^19 C and mass 1.67×10í^27
kg) that is placed in this field?
Answer:a = m/s^2