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Chapter 20: Problem 23

A 68 -nC charge experiences a 150 -mN force in a certain electric field. Find (a) the field strength and (b) the force that a \(35-\mu C\) charge would experience in the same field.

Short Answer

Expert verified
The field strength is \( 2.21 * 10^6 N/C\) and the force that a \(35-\mu C\) charge would experience in the same field is \(77.35 N\)

Step by step solution

01

Calculate the electric field

Begin by arranging the electric field strength formula to isolate E: \(E=F/q\). Then, substitute the known values into the formula. It's important to note that we need to convert the charge from nC to C, since our force is given in N. Hence, \(68nC = 68 * 10^-9 C\). Therefore: \(E=(150 * 10^-3 N) / (68*10^{-9} C)= 2.21 * 10^6 N/C\)
02

Calculate the force on a different charge

Next, use the calculated electric field strength and the formula for electric force \(F=qE\) to find the force on a different charge. Again note the units: this time convert \(35 - \mu C\) to \(35 * 10^-6 C\). So: \(F=(35*10^{-6} C)*(2.21 * 10^6 N/C)= 77.35 N\)

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Most popular questions from this chapter

A charge \(q\) is at the point \(x=1 \mathrm{m}, y=0 \mathrm{m} .\) Write expressions for the unit vectors you would use in Coulomb's law if you were finding the force that \(q\) exerts on other charges located at (a) \(x=1 \mathrm{m}, y=1 \mathrm{m} ;\) (b) the origin; and \((\mathrm{c}) x=2 \mathrm{m}, y=3 \mathrm{m}\) You're not given the sign of \(q .\) Why doesn't this matter?

A \(1.0-\mu C\) charge and a \(2.0-\mu \mathrm{C}\) charge are \(10 \mathrm{cm}\) apart. Find a point where the electric field is zero.

Under what circumstances is the path of a charged particle a parabola? A circle?

A 2 -g ping-pong ball rubbed against a wool jacket acquires a net positive charge of \(1 \mu \mathrm{C}\). Estimate the fraction of the ball's electrons that have been removed.

A thin rod extends along the \(x\) -axis from \(x=0\) to \(x=L\) and carries line charge density \(\lambda=\lambda_{0}(x / L)^{2},\) where \(\lambda_{0}\) is a constant. Find the electric field at \(x=-L\).
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