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In a salt crystal, the distance between adjacent sodium and chloride ions is 2.82 × 10 −10 m . What is the force of attraction between the two singly charged ions?

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Protons in an atomic nucleus are typically 10 −15 m apart. What is the electric force of repulsion between nuclear protons?

F = 230.7 N

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Suppose Earth and the Moon each carried a net negative charge − Q . Approximate both bodies as point masses and point charges.

(a) What value of Q is required to balance the gravitational attraction between Earth and the Moon?

(b) Does the distance between Earth and the Moon affect your answer? Explain.

(c) How many electrons would be needed to produce this charge?

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Point charges q 1 = 50 μ C and q 2 = −25 μ C are placed 1.0 m apart. What is the force on a third charge q 3 = 20 μ C placed midway between q 1 and q 2 ?

F = 53.94 N

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Where must q 3 of the preceding problem be placed so that the net force on it is zero?

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Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge Q , the threads hang at 5.0 ° to the vertical, as shown below. What is the magnitude of Q ? What are the signs of the two charges?

Two small balls are attached to threads which are in turn tied to the same point on the ceiling. The threads hang at an angle of 5.0 degrees to either side of the vertical. Each ball has a charge Q.

The tension is T = 0.049 N . The horizontal component of the tension is 0.0043 N
d = 0.088 m , q = 6.1 × 10 −8 C .
The charges can be positive or negative, but both have to be the same sign.

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Point charges Q 1 = 2.0 μ C and Q 2 = 4.0 μ C are located at r 1 = ( 4.0 i ^ 2.0 j ^ + 5.0 k ^ ) m and r 2 = ( 8.0 i ^ + 5.0 j ^ 9.0 k ^ ) m . What is the force of Q 2 on Q 1 ?

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The net excess charge on two small spheres (small enough to be treated as point charges) is Q . Show that the force of repulsion between the spheres is greatest when each sphere has an excess charge Q /2. Assume that the distance between the spheres is so large compared with their radii that the spheres can be treated as point charges.

Let the charge on one of the spheres be rQ , where r is a fraction between 0 and 1. In the numerator of Coulomb’s law, the term involving the charges is r Q ( 1 r ) Q . This is equal to ( r r 2 ) Q 2 . Finding the maximum of this term gives 1 2 r = 0 r = 1 2

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Two small, identical conducting spheres repel each other with a force of 0.050 N when they are 0.25 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.060 N. What is the original charge on each sphere?

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A charge q = 2.0 μ C is placed at the point P shown below. What is the force on q ?

Two charges are shown, placed on a horizontal line and separated by 2.0 meters. The charge on the left is a positive 1.0 micro Coulomb charge. The charge on the right is a negative 2.0 micro Coulomb charge. Point P is 1.0 to the right of the negative charge.

Define right to be the positive direction and hence left is the negative direction, then F = −0.05 N

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What is the net electric force on the charge located at the lower right-hand corner of the triangle shown here?

Charges are shown at the vertices of an equilateral triangle with sides length a. The bottom of the triangle is on the x axis of an x y coordinate system, and the bottom left vertex is at the origin. The charge at the origin is positive q. The charge at the bottom right hand corner is also positive q. The charge at the top vertex is negative two q.
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Two fixed particles, each of charge 5.0 × 10 −6 C , are 24 cm apart. What force do they exert on a third particle of charge −2.5 × 10 −6 C that is 13 cm from each of them?

The particles form triangle of sides 13, 13, and 24 cm. The x -components cancel, whereas there is a contribution to the y -component from both charges 24 cm apart. The y -axis passing through the third charge bisects the 24-cm line, creating two right triangles of sides 5, 12, and 13 cm.
F y = 2.56 N in the negative y -direction since the force is attractive. The net force from both charges is F net = −5.12 N j ^ .

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The charges q 1 = 2.0 × 10 −7 C, q 2 = −4.0 × 10 −7 C, and q 3 = −1.0 × 10 −7 C are placed at the corners of the triangle shown below. What is the force on q 1 ?

Charges are shown at the vertices of a right triangle. The bottom of the triangle is length 4 meters, the vertical side on the left is length 3 meters, and the hypotenuse is length 5 meters. The charge at the top is q sub one and positive, the charge at the bottom left is q sub 3 and negative and the charge at the bottom right is q sub 2 and negative.
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What is the force on the charge q at the lower-right-hand corner of the square shown here?

Charges are shown at the corners of a square with sides length a. All of the charges are positive and all are magnitude q.

The diagonal is 2 a and the components of the force due to the diagonal charge has a factor cos θ = 1 2 ;
F net = [ k q 2 a 2 + k q 2 2 a 2 1 2 ] i ^ [ k q 2 a 2 + k q 2 2 a 2 1 2 ] j ^

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Point charges q 1 = 10 μ C and q 2 = −30 μ C are fixed at r 1 = ( 3.0 i ^ 4.0 j ^ ) m and r 2 = ( 9.0 i ^ + 6.0 j ^ ) m . What is the force of q 2 on q 1 ?

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Practice Key Terms 6

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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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