BIO Base Pairing in DNA, II. Refer to Exercise 21.21. Figure E21.22 shows the bonding of cytosine and guanine. The O—H and H—N distances are each 0.110 nm. In this case, assume that the bonding is due only to the forces along the O—H—O, N—H—N, and O—H—N combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive? Figure E21.22
BIO Base Pairing in DNA, II. Refer to Exercise 21.21. Figure E21.22 shows the bonding of cytosine and guanine. The O—H and H—N distances are each 0.110 nm. In this case, assume that the bonding is due only to the forces along the O—H—O, N—H—N, and O—H—N combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive? Figure E21.22
BIO Base Pairing in DNA, II. Refer to Exercise 21.21. Figure E21.22 shows the bonding of cytosine and guanine. The O—H and H—N distances are each 0.110 nm. In this case, assume that the bonding is due only to the forces along the O—H—O, N—H—N, and O—H—N combinations, and assume also that these three combinations are parallel to each other. Calculate the net force that cytosine exerts on guanine due to the preceding three combinations. Is this force attractive or repulsive?
8
What is the value of the electrical force acting on the object according to the system in the figure? (Surface charge density is 8.85.10^(-9) C/m^2, q= 4.10^(-3)C)
a)
1,2
B)
one thousand
NS)
5.6
D)
one
TO)
4
Use the following constants if necessary. Coulomb constant, k=8.987×109N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10−12F/m. The magnitude of the Charge of one electron, e=−1.60217662×10−19C. Mass of one electron, me=9.10938356×10−31kg. Unless specified otherwise, each symbol carries its usual meaning. For example, μC means microcoulomb.
A line of charge of length L=12m with charge density λ=48.0μC/m lies along the positive Y axis whose one end is at the origin O. A point charge q=49.0μC lies on the X axis at (44,0,0) and point P lies on the Z axis at 13m from the origin. Here the coordiates are given in meters.
Given, the electric field at P due to only the point charge q:
x - component = -200.6
y - component = 0
z - component = 59.3
a) Find the x, y and z components of the electric field at P due to the line of charge only. (Hint: You may have to do an integration.)
b) Find the x, y and z components of the net electric field at point P
Use the following constants if necessary. Coulomb constant, k=8.987×10^9N⋅m2/C2. Vacuum permittivity, ϵ0=8.854×10^−12F/m. Magnitude of the Charge of one electron, e=−1.60217662×10^−19C. Mass of one electron, me=9.10938356×10^−31kg. Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb.
Suppose you have 5-point charges q1 to q5 distributed along a semi-circle maintaining equal distance. The charge q6 is on the center of the circle which is also the origin of given coordinate system. The radius of the semi-circle is R=18cm and the magnitude of the charges are as follows: q1=q5=33μC, q2=q4=27μC and q3=q6=−33μC.
a)Find the vector that points from q1to q6 and call it r⃗ 1,6.
x component of the vector
y component of the vector b) Calculate the coulomb force on q6 due to q1 in unit vector notation and call it F⃗ 1,6.
x component of the force
y component of the force
c)Find the vector that points from q3to q6 and call it r⃗…
Chapter 21 Solutions
University Physics with Modern Physics (14th Edition)
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