Skip to main content
Ch 23: Electric Potential

Chapter 23, Problem 23

At a certain distance from a point charge, the poten-tial and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m, respectively. (Take V = 0 at infinity.) (a) What is the distance to the point charge?

Verified Solution
Video duration:
5m
This video solution was recommended by our tutors as helpful for the problem above.
1263
views
Was this helpful?

Video transcript

Hey everyone welcome back in this problem. We have an electric potential at point P from a charged particle. That is 10.2 volts. The electric field at that same point has a magnitude of 22. volts per meter. And we are asked to calculate the length from the charge two point Okay. And we're told to assume that the potential is zero at infinity. So we're giving information about electric potential and about the electric field. So let's recall that we can write the electric potential V Equal to 1/4 Pi. Absolutely not Q over R. Okay. We're talking about just one charged protocol. So we just have to deal with um one term here and we can write the magnitude of the electric field E as 1/4 pi epsilon. Not absolute value of Q divided by R squared. Okay. Alright. Let's fill in the information that we know and figure out what we want to find. Okay, well we know that the electric potential is 10.2 volts. Okay. On the left hand side here we have 10.2 volts is equal to 1/4 pi epsilon, not Q. Which we don't know divided by R. Which we also don't know. But our is what we're trying to find the length from the charge to the point P. So we're trying to find that distance R. Alright, let's do the same for our equation for the electric field. Ok. We're told that the electric field has a magnitude of 22.5 volts per meter. So we get 22.5 volts per meter is equal to 1/4 pi epsilon. Not whoops. Not divided by times absolute value of Q divided by R squared. Okay. And again this is the same R as in the other equation that we're trying to solve for. And what you'll notice is that we have two equations and two unknowns. Okay, We have our equation for the electric potential. We have our equation for the electric field and we have Q. And are those two values that we don't know? We don't We are asked to find information about Q. We are only asked to find information about our Okay, so let's start by isolating Q in our first equation. Okay. We're gonna write queue in terms of our then we can substitute that into the other equation and we'll have an equation just in terms of our we can solve for that our value. Okay? So if we write the left hand equation for the potential in terms of Q, we're gonna have is Q. Is equal to Okay. And we multiply it. So we get 10.2 volts times four pi epsilon naught times are no we're gonna take this we'll call this star in blue. We're gonna substitute that into our equation for the electric field. Okay? So if we substitute this information, we're gonna have 22.5 volts per meter. It's equal to 1/4 pi epsilon. Not the absolute value of Q. Which is now going to be 10.2 volts times four pi epsilon. Not times the absolute value of our Okay. And this is all divided by our squirt. Okay, so we figured out what Q was in terms of our and our first equation. We substituted that into the second equation and now we just have an equation with ours and numbers we can go ahead and solve for R. And you'll notice that those four pi epsilon naught cancels here And we're left with 22.5V/m is equal to. We have 10.2V. We have absolute value of our divided by R squared. This is going to give us the absolute value of our and the denominator. Okay, solving for our give ourselves a bit more room And we find that the absolute value of R is equal to 10.2V divided by 22. volts per meter. Okay, volt divided by volts per meter is gonna give us a unit of meter and we get 0.453 m. Okay? And so that is gonna be the length Between that charge and the point of interest. 0.453 m. And if we go back up to our answers, we see that that is going to correspond with answer. Thanks everyone for watching. I hope this video helped see you in the next one
Related Practice
Textbook Question
Two point charges of equal magnitude Q are held a distance d apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?).
1484
views
3
rank
Textbook Question
Two point charges q_1 = +2.40 nC and q_2 = -6.50 nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q_1 and 0.060 m from q_2 (Fig. E23.19). Take the electric potential to be zero at infinity. Find (a) the potential at point A.

1200
views
1
rank
Textbook Question
Two point charges q_1 = +2.40 nC and q_2 = -6.50 nC are 0.100 m apart. Point A is midway between them; point B is 0.080 m from q_1 and 0.060 m from q_2 (Fig. E23.19). Take the electric potential to be zero at infinity. Find (b) the potential at point B.

443
views
Textbook Question
An electron is to be accelerated from 3.00x10^6 m/s to 8.00x10^6 m/s. (a) Through what potential difference must the electron pass to accomplish this?
832
views
1
rank
Textbook Question
At a certain distance from a point charge, the poten-tial and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m, respectively. (Take V = 0 at infinity.) (b) What is the magnitude of the charge?
449
views
1
rank
Textbook Question
At a certain distance from a point charge, the poten-tial and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m, respectively. (Take V = 0 at infinity.) (c) Is the electric field directed toward or away from the point charge?
625
views