The work done by an external force to move a - 7.40 μC charge ...

Question

The work done by an external force to move a - 7.40 μC charge from point A to point B is 15.0 x 10^-4 J. If the charge was started from rest and had 4.82 x 10^-4 J of kinetic energy when it reached point B, what must be the potential difference between A and B?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem, consider that the external force moves a stationary charge of negative 8.0 micro Coolum from point P to point Q. The work done during this process is 18 multiplied by 10 to the power of negative four joules determine the voltage difference between points P and Q. If the charge gained a kinetic energy of about 7.0 multiplied by 10 to the power of negative four joules upon reaching point Q. So that's our goal. Our angles are trying to figure out what the voltage differences between points P and Q provided the conditions given to us by the problem itself. So now that we know that we're ultimately solving for the voltage difference between points P and Q, let's read off our multiple choice answers to see what our final answer might be noting that they're all in the same units of volts. So A is negative 1.4 multiplied by 10 to the power of two B is negative 1.3 multiplied by 10 to the power of two C is 1.3 multiplied by 10 to the power of two and D is 1.4 multiplied by 10 to the power of two. OK. So first off, let us write down all of our known variables. So first off, let us note that we know that the charge, which we're gonna denote sq we know that the charge Q is equal to negative eight micro cools. And we know that eight, negative eight micro Coolum is also equal to negative eight multiplied by 10 to the power of negative six forms. I've just gone ahead and given you the conversion, but you can use dimensional analysis to solver this or you just quickly look it up. And we also know that the work done which we're gonna call W and this is the work done by the external course, is equal to 18 multiplied by 10 to the power of negative four jewels. And we also know the kinetic energy at point so that they're saying that the, we also know that the charge gained a kinetic upon reaching Q. So we know that the kinetic energy at point Q is, and let's call this keq and keq is equal to 7.0 multiplied by 10 to the power of negative four jewels. Awesome. So now our second step is we need to figure out the relationship between the work kinetic energy and the electric potential energy by recalling and using our work equation which states that the work is equal to the change in kinetic energy. So delta ke plus delta U which is the change in potential energy. So now our third step is we need to figure out the change in kinetic energy will be the kinetic energy from at Q minus the kinetic energy at P. So at point Q from point P, so let's make a little note of that off to the side here. So K A EQ Q minus ke don't forget your kkep. Awesome. So with that in mind, hold on to that thought. And we also need to note and recall that since the initial kinetic energy kep was zero, we can go ahead and write that delta K is equal to seven 0.0. So 7.0 multiplied by 10 to the power of negative four joules. And this is because let's make a little note here that Kep is equal to zero. Awesome. So our fourth step is, is we need to figure out the change in electric potential energy by recalling and using the following equation that delta U is equal to W minus delta ke which is equal to when we plug in our known variables, 18 multiplied by 10 to the power of negative four joules for work minus our change in kinetic energy which is 7.07 0.0 multiplied by 10 to the power of negative four jules, which will mean that delta U is equal to 0.0011 or 1.1 multiplied by 10 to the power of negative three jewels. Awesome. So now our final step is we need to figure out the voltage, we're trying to figure out what the voltage difference is. So we could calculate the voltage difference by recalling and using the ball equation. So we're gonna say V the voltage difference is equal to delta U divided by QR charge, which when we plug in our known variables, we will get that delta U which we just solve together is equal to 1.1 multiplied by 10 to the power of negative three jewels divided by our charge Q which is negative eight multiplied by 10 to the power of negative six cool lumps which we plug that into our calculator and we round to one decimal place. Writing our final answer in scientific notation, we will get negative 1.4 multiplied by 10 to the power of two volts. And that's it. That's our final answer. Hooray. We did it. So looking at our multiple choice answers, the correct answer has to be the letter A negative 1.4 multiplied by 10 to the power of two volts. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

The work done by an external force to move a - 7.40 μC charge from point A to point B is 15.0 x 10^-4 J. If the charge was started from rest and had 4.82 x 10^-4 J of kinetic energy when it reached point B, what must be the potential difference between A and B?

Key Concepts

Work-Energy Principle

Electric Potential Difference

Conservation of Energy

Watch next

The work done by an external force to move a - 7.40 μC charge from point A to point B is 15.0 x 10-4 J. If the charge was started from rest and had 4.82 x 10-4 J of kinetic energy when it reached point B, what must be the potential difference between A and B?

Key Concepts

Work-Energy Principle

Electric Potential Difference

Conservation of Energy

Watch next

The work done by an external force to move a - 7.40 μC charge from point A to point B is 15.0 x 10^-4 J. If the charge was started from rest and had 4.82 x 10^-4 J of kinetic energy when it reached point B, what must be the potential difference between A and B?