Calculating work done by an Electric field on a positive charge

[psy]

Hey

How can i calculate the work done
by a homogeneous electric field on a positive charge q
= 7 × 10^-8 C when the charge moves from point 1 to point 2
. I have two cases:
a) the charge is parallel to the
Electric field;
b) the displacement of the charge takes place at the angle α = 60 °
to the
Electric field.
The electric field strength is E = 6 × 105 V / m, the displacement
of the charge is s = 10 cm

https://www.flickr.com/photos/155324944@N02/35099011562/in/dateposted/
So the mechanical work equation is equivalent to the electrical

case a)
The force done on the particle is the electric field strength multiplied with its charge :

F = q * E = 7* 10^-8 C * 6 * 10^5 V/m = 42 * 10^-3 CV/m

The work is given by W = F*s = 42 * 10^-3 CV/m * 0.1 m = 4.2 *10^-3 CV ( m*kg*A^2 / s^2)

case b)
the particle is still moving in the same direction as the field force ,but a with an angle,
so i would use

F = q * E * cos(60°)

Is this a proper way to calculate it?

 

[Doc Al]

The first case (a) looks good to me. (The standard unit for work is the Joule.)

 

[ehild]

case b)
the particle is still moving in the same direction as the field force ,but a with an angle,
so i would use

F = q * E * cos(60°)

No, the equation is wrong. The electric force is proportional to the electric field, $\vec F = q \vec E$
The problem text says that "b) the displacement of the charge takes place at the angle α = 60 °to the Electric field."
The charge is scalar, it does not have direction.
What is the work if the force and the displacement make an angle α ?

 

[Doc Al]

case b)
the particle is still moving in the same direction as the field force ,but a with an angle,
so i would use

F = q * E * cos(60°)

As ehild points out, that equation is incorrect. But it's a step in the right direction if you meant to give the component of the force in the direction of the displacement.

 

[gneill]

The electric field strength is E = 6 × 105 V / m, the displacement
of the charge is s = 10 cm

https://www.flickr.com/photos/155324944@N02/35099011562/in/dateposted/

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