Electric field question

[erucibon]

Points P and Q are 10mm apart in an electric field. When a charge of +8.0C is moved from P to Q, the field does 4x10^-5 J of work. What is the direction of the field and why.

Thanks

[DrDusk]

Points P and Q are 10mm apart in an electric field. When a charge of +8.0C is moved from P to Q, the field does 4x10^-5 J of work. What is the direction of the field and why.

Thanks

Instead of giving you the answer, I'm going to give you a very similar scenario which should allow you to figure this out.

Consider this scenario. Your on Earth and you throw a ball up. Work done is the change in Kinetic energy if you recall. This means we can write the work done for the ball as



Now if you look at the equation, you can see that it will be positive if v_f is greater than v_i and negative if v_f is less than v_i. Now if the ball is thrown up, we know its velocity is decreasing as it goes up. This means v_f will always be less than v_i when it's going up. Now if gravity accelerates objects in the direction it points, we can easily figure out that gravity must be pointing downwards, because if it was point upwards, v_f would be greater than v_i, meaning the Work done would be positive, but we know v_f is less than v_i as the ball is slowing down.

Apply the same concept for the electric field one, however you will need to consider the direction in which a positive charge will move under an Electric field, i.e. will it move against the field or in the direction of the field...

[erucibon]

Instead of giving you the answer, I'm going to give you a very similar scenario which should allow you to figure this out.

Consider this scenario. Your on Earth and you throw a ball up. Work done is the change in Kinetic energy if you recall. This means we can write the work done for the ball as



Now if you look at the equation, you can see that it will be positive if v_f is greater than v_i and negative if v_f is less than v_i. Now if the ball is thrown up, we know its velocity is decreasing as it goes up. This means v_f will always be less than v_i when it's going up. Now if gravity accelerates objects in the direction it points, we can easily figure out that gravity must be pointing downwards, because if it was point upwards, v_f would be greater than v_i, meaning the Work done would be positive, but we know v_f is less than v_i as the ball is slowing down.

Apply the same concept for the electric field one, however you will need to consider the direction in which a positive charge will move under an Electric field, i.e. will it move against the field or in the direction of the field...

Is it because the since the work done is positive and W = change in potential, Uf > Ui. As a charge moves from high to low potential, and a postive charge is in the same direction as the field, the field is moving from Q to P. But does this mean that there is an external force greater than the field being applied so that it is able to move from P to Q?

[DrDusk]

Why are you thinking of work in terms of change in potential? Think of it in terms of Kinetic Energy

Also Work is (-1 * change in potential), not the change in potential.

And that is incorrect. The field should be going from P to Q