The field due to two parallel plates, the left with a negative surface charge density of −6.0 C/m2 whilethe sheet on the right has a positive surface charge density of +9.0 C/m2is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss’s law that, independent of this distance,• In I the net field points to the left and has magnitude |Enet| = 1.69 × 1011 N/C.• In region II, between the plates, the field has magnitude 8.47×1011 N/C and points to the left• In III it points to the right and has magnitude |Enet| = 1.69 × 1011 N/C. In I the net field points to the left and has magnitude |Enet| = 1,69 × 10¹¹ N/C.• In region II, between the plates, the field has magnitude 8.47 × 10¹¹ N/C and points to the left• In III it points to the right and has magnitude |Enet| = 1,69 × 10¹¹ N/C.·|·|III(a) Plot qualitatively correct E(x) and V(x) graphs, setting V=0 at the left (negatively charged)plate and letting x be the distance from the left plate, with positive x to the right. Use positivevalues for E if it points to the right and negative if it points to the left. Your plot should includethe range of x from -0.002 m to +0.004 m. Check that the direction and size of the slope ofyour V graph is consistent with your E graph values.(b) In what direction would a positive test charge be pushed in each region? Discuss from the pointof view of field and from the point of view of voltage.(c) What would happen in each region to a negative test charge moving to the right? (Would itsspeed increase, decrease, remain the same?)(d) How would your graph change if you chose V=0 to be at the positive plate? Draw that graph.(e) What is the potential difference (AV) between the plates?

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Answered: • In I the net field points to the left…

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter25: Electric Potential

Section: Chapter Questions

Problem 25.56AP: Review. From a large distance away, a particle of mass m1, and positive charge q1 is fired at speed ...

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The field due to two parallel plates, the left with a negative surface charge density of −6.0 C/m2 while
the sheet on the right has a positive surface charge density of +9.0 C/m2
is given below. Suppose the plates were spaced 0.0020 m apart. We found that, using Gauss’s law that, independent of this distance,

• In I the net field points to the left and has magnitude |Enet| = 1.69 × 1011 N/C.
• In region II, between the plates, the field has magnitude 8.47×1011 N/C and points to the left
• In III it points to the right and has magnitude |Enet| = 1.69 × 1011 N/C.

In I the net field points to the left and has magnitude |Enet| = 1,69 × 10¹¹ N/C.
• In region II, between the plates, the field has magnitude 8.47 × 10¹¹ N/C and points to the left
• In III it points to the right and has magnitude |Enet| = 1,69 × 10¹¹ N/C.
·|·|
III
(a) Plot qualitatively correct E(x) and V(x) graphs, setting V=0 at the left (negatively charged)
plate and letting x be the distance from the left plate, with positive x to the right. Use positive
values for E if it points to the right and negative if it points to the left. Your plot should include
the range of x from -0.002 m to +0.004 m. Check that the direction and size of the slope of
your V graph is consistent with your E graph values.
(b) In what direction would a positive test charge be pushed in each region? Discuss from the point
of view of field and from the point of view of voltage.
(c) What would happen in each region to a negative test charge moving to the right? (Would its
speed increase, decrease, remain the same?)
(d) How would your graph change if you chose V=0 to be at the positive plate? Draw that graph.
(e) What is the potential difference (AV) between the plates?

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