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What is the force on the 2.0 - μ C charge placed at the center of the square shown below?

Charges are shown at the corners of a square with sides length 1 meter. The top left charge is positive 5.0 micro Coulombs. The top right charge is positive 4.0 micro Coulombs. The bottom left charge is negative 4.0 micro Coulombs. The bottom right charge is positive 2.0 micro Coulombs. A fifth charge of positive 2.0 micro Coulombs is at the center of the square.
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Four charged particles are positioned at the corners of a parallelogram as shown below. If q = 5.0 μ C and Q = 8.0 μ C , what is the net force on q ?

Four charges are positioned at the corners of a parallelogram. The top and bottom of the parallelogram are horizontal and are 3.0 meters long. The sides are at a thirty degree angle to the x axis. The vertical height of the parallelogram is 1.0 meter. The charges are a positive Q in the lower left corner, positive 2 Q in the lower right corner, negative 3 Q in the upper left corner, and positive q in the upper right corner.

Charges Q and q form a right triangle of sides 1 m and 3 + 3 m . Charges 2 Q and q form a right triangle of sides 1 m and 3 m .
F x = 0.036 N ,
F y = 0.09 N ,
F net = 0.036 N i ^ + 0.09 N j ^

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A charge Q is fixed at the origin and a second charge q moves along the x -axis, as shown below. How much work is done on q by the electric force when q moves from x 1 to x 2 ?

A charge Q is shown at the origin and a second charge q is shown to its right, on the x axis, moving to the right. Both are positive charges. Point x 1 is between the charges. Point x 2 is to the right of both.
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A charge q = −2.0 μ C is released from rest when it is 2.0 m from a fixed charge Q = 6.0 μ C . What is the kinetic energy of q when it is 1.0 m from Q ?

W = 0.054 J

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What is the electric field at the midpoint M of the hypotenuse of the triangle shown below?

Charges are shown at the vertices of an isosceles right triangle whose sides are length a and those hypotenuse is length M. The right angle is the bottom right corner. The charge at the right angle is positive 2 q. Both of the other two charges are positive q.
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Find the electric field at P for the charge configurations shown below.

In figure a, positive charge q is on the left, negative charge q is a distance a to the right of it. Point P is a distance a to the right of the negative charge q. In figure b, positive charge q is on the left, and a positive charge q is a distance a to the right of it. Point P is below the midpoint, a distance a from each of the charges so that the two charges and point P are at the vertices of an equilateral triangle whose sides are length a. In figure c, four charges are at the corners of a square whose sides are length a. The two top corners each have positive charge q. The two bottom corners each have negative charge q. Point P is at the center of the square.

a. E = 1 4 π ε 0 ( q ( 2 a ) 2 q a 2 ) i ^ ; b. E = 3 4 π ε 0 q a 2 ( j ^ ) ; c. E = 2 π ε 0 q a 2 1 2 ( j ^ )

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(a) What is the electric field at the lower-right-hand corner of the square shown below? (b) What is the force on a charge q placed at that point?

A square with sides of length a is shown. Three charges are shown as follows: At the top left, a charge of negative 2 q. At the top right, a charge of positive q. At the lower left, a charge of positive q.
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Point charges are placed at the four corners of a rectangle as shown below: q 1 = 2.0 × 10 −6 C , q 2 = −2.0 × 10 −6 C, q 3 = 4.0 × 10 −6 C, and q 4 = 1.0 × 10 −6 C . What is the electric field at P ?

A rectangle is shown with a charge at each corner. The rectangle is 4.0 centimeters high and 6.0 centimeters wide. At the top left is a positive charge q 1. At the top right is a negative charge q 2. At the lower left is a positive charge q 3. At the lower right is a positive charge q 4. Point P is in the middle of the upper edge, 3.0 centimeters to the right of q 1 and 3.0 centimeters to the left of q 2.

E = 6.4 × 10 6 ( i ^ ) + 1.5 × 10 7 ( j ^ ) N / C

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Three charges are positioned at the corners of a parallelogram as shown below. (a) If Q = 8.0 μ C , what is the electric field at the unoccupied corner? (b) What is the force on a 5.0 - μ C charge placed at this corner?

Three charges are positioned at the corners of a parallelogram. The top and bottom of the parallelogram are horizontal and are 3.0 meters long. The sides are at a thirty degree angle to the x axis. The vertical height of the parallelogram is 1.0 meter. The charges are a positive Q in the lower left corner, positive 2 Q in the lower right corner, and negative 3 Q in the upper left corner.
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A positive charge q is released from rest at the origin of a rectangular coordinate system and moves under the influence of the electric field E = E 0 ( 1 + x / a ) i ^ . What is the kinetic energy of q when it passes through x = 3 a ?

F = q E 0 ( 1 + x / a ) W = 1 2 m ( v 2 v 0 2 ) ,
1 2 m v 2 = q E 0 ( 15 a 2 ) J

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A particle of charge q and mass m is placed at the center of a uniformaly charged ring of total charge Q and radius R . The particle is displaced a small distance along the axis perpendicular to the plane of the ring and released. Assuming that the particle is constrained to move along the axis, show that the particle oscillates in simple harmonic motion with a frequency f = 1 2 π q Q 4 π ε 0 m R 3 .

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Charge is distributed uniformly along the entire y -axis with a density λ y and along the positive x -axis from x = a to x = b with a density λ x . What is the force between the two distributions?

Electric field of wire at x : E ( x ) = 1 4 π ε 0 2 λ y x i ^ ,
d F = λ y λ x 2 π ε 0 ( ln b ln a )

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The circular arc shown below carries a charge per unit length λ = λ 0 cos θ , where θ is measured from the x -axis. What is the electric field at the origin?

An arc that is part of a circle of radius r and with center at the origin of an x y coordinate system is shown. The arc extends from an angle theta sub zero above the x axis to an angle theta sub zero below the x axis.
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Calculate the electric field due to a uniformly charged rod of length L , aligned with the x -axis with one end at the origin; at a point P on the z -axis.


A rod of length L is shown, aligned with the x-axis with the left end at the origin. A point P is shown on the z axis, a distance a above the left end of the rod. A small segment of the rod is labeled as d x and is a distance x to the right of the left end of the rod. The line from dx to point P makes an angle of theta with the x axis. The vector d E, drawn with its tail at point P, points away from the segment d x.
d E x = 1 4 π ε 0 λ d x ( x 2 + a 2 ) x x 2 + a 2 ,
E x = λ 4 π ε 0 [ 1 L 2 + a 2 1 a ] i ^ ,
d E z = 1 4 π ε 0 λ d x ( x 2 + a 2 ) a x 2 + a 2 ,
E z = λ 4 π ε 0 a L L 2 + a 2 k ^ ,
Substituting z for a , we have:
E ( z ) = λ 4 π ε 0 [ 1 L 2 + z 2 1 z ] i ^ + λ 4 π ε 0 z L L 2 + z 2 k ^

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The charge per unit length on the thin rod shown below is λ . What is the electric force on the point charge q ? Solve this problem by first considering the electric force d F on q due to a small segment d x of the rod, which contains charge λ d x . Then, find the net force by integrating d F over the length of the rod.

A rod of length l is shown. The rod lies on the horizontal axis, with its left end at the origin. A positive charge q is on the x axis, a distance a to the right of the right end of the rod.
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The charge per unit length on the thin rod shown here is λ . What is the electric force on the point charge q ? (See the preceding problem.)

A rod of length l is shown. The rod lies on the horizontal axis, with its center at the origin, so the ends are a distance of l over 2 to the left and right of the origin. A positive charge q is on the y axis, a distance a to above the origin.

There is a net force only in the y -direction. Let θ be the angle the vector from dx to q makes with the x -axis. The components along the x -axis cancel due to symmetry, leaving the y -component of the force.
d F y = 1 4 π ε 0 a q λ d x ( x 2 + a 2 ) 3 / 2 ,
F y = 1 2 π ε 0 q λ a [ l / 2 ( ( l / 2 ) 2 + a 2 ) 1 / 2 ]

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The charge per unit length on the thin semicircular wire shown below is λ . What is the electric force on the point charge q ? (See the preceding problems.)

A semicircular arc that the upper half of a circle of radius R is shown. A positive charge q is at the center of the circle.
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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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