Home Page Textbooks Physics Page 743: Standardized Test Prep

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 743: Standardized Test Prep

Exercise 1

Step 1

1 of 7

The magnitude of an electric field due to a point charge is given by the equation:
$$begin{align*}
E=kfrac{vert q vert}{r^2}
end{align*}$$
where $k=8.99times 10^9;text{N}cdottext{m}^2/text{C}^2$, and $r$ is the distance of the field from the point charge $q$.

Step 2

2 of 7

We let $q_1=2;mu$C at $x=1$ m. Since the electric field of interest is at $x=3$ m, the distance of the field from the point charge $q_1$ is 2 m. Substituting these values to the equation, we get:
$$begin{align*}
E_1=(8.99times 10^9;text{N}cdottext{m}^2/text{C}^2)frac{vert 2times10^{-6};text{C} vert}{(2;text{m})^2}=4495;frac{text{N}}{text{C}}
end{align*}$$

Step 3

3 of 7

Note that the direction of an electric field is toward a negative charge and away from a positive charge. Since $q_1$ is a positive charge, this would mean that the electric field will point along the positive direction of $x$.
$$begin{align*}
vec{E_1}=4495;frac{text{N}}{text{C}};hat{x}
end{align*}$$

Step 4

4 of 7

We repeat the same procedure for the other point charge which is $q_2=-4;mu$C at $x=6$ m. The distance of the field from this point charge is 3 m. Substituting our known values, we get:
$$begin{align*}
E_2=(8.99times 10^9;text{N}cdottext{m}^2/text{C}^2)frac{vert -4times10^{-6};text{C} vert}{(3;text{m})^2}=3996;frac{text{N}}{text{C}}
end{align*}$$

Step 5

5 of 7

$q_2$ is a negative charge; thus, the electric field will point towards $q_2$ which is along the positive $x$ direction.
$$begin{align*}
vec{E_2}=3996;frac{text{N}}{text{C}};hat{x}
end{align*}$$

Step 6

6 of 7

The net electric field is obtained through the vector sum of the electric fields due to the individual charges.
$$begin{align*}
vec{E_{net}} &=vec{E_1}+vec{E_2}\
vec{E_{net}} &=4495;frac{text{N}}{text{C}};hat{x}+3996;frac{text{N}}{text{C}};hat{x}\
vec{E_{net}} &= 8491;frac{text{N}}{text{C}};hat{x}\
vec{E_{net}} &approx boxed{8500;frac{text{N}}{text{C}};hat{x}}
end{align*}$$

Result

7 of 7

$text{A}$.

Exercise 2

Step 1

1 of 6

The net electric field due to two charges is given by the vector sum of the electric fields due to the individual charges.
$$begin{align*}
vec{E_{net}}=vec{E_1}+vec{E_2}
end{align*}$$

Step 2

2 of 6

At some point between the charges, the vector fields due to each charge are in the same direction which is in the positive $x$ direction. Therefore would not cancel out and produce a net electric field that is nonzero.

Step 3

3 of 6

At some point to the right of $x=6$ m, the vector fields from the two charges are in the opposite direction with each other which is needed to cancel them out. However, charge $q_2=-4;mutext{C}$, the larger charge, will generate a smaller distance $r$ to this point and charge $q_1=2;mutext{C}$, the smaller charge, will have a larger $r$ distance to the point. The equation for the magnitude of the electric field due to a point charge is given by:
$$begin{align*}
E=kfrac{vert q vert}{r^2}
end{align*}$$

Step 4

4 of 6

With this, a larger magnitude of electric field due to $q_2$ and a smaller magnitude due to $q_1$ would be produced and therefore would not cancel each other out.

Step 5

5 of 6

At some point to the left of $x=1$ m, the vector fields are also opposite in direction with another. Also, the smaller charge $q_1$ would generate a smaller distance $r$ to this point and the larger charge $q_2$ would generate a larger distance $r$ to this point. With that, there would be a point where their magnitudes would be the same;therefore, they would cancel out. Thus, $textbf{the net electric field would be zero at some point to the left of}$ $mathbf{x=1}$ $textbf{m.}$

Result

6 of 6

$text{B.}$

Exercise 3

Step 1

1 of 4

The magnitude of the electric field due to a point charge $q$ at a distance $r$ is given by:
$$begin{align*}
E=kfrac{vert q vert}{r^2}
end{align*}$$

Step 2

2 of 4

It is given in the problem that $E$ is the field due to an isolated charge $Q$ at a distance $d$. Thus,
$$begin{align*}
E=kfrac{vert Q vert}{d^2}
end{align*}$$

Step 3

3 of 4

We let $E’$ be the new magnitude of the electric field. Doubling the charge to $2Q$ and the distance to $2d$, we get this relationship:
$$begin{align*}
E’=kfrac{vert 2Q vert}{(2d)^2}=kfrac{vert 2Q vert}{4d^2}=kfrac{vert Q vert}{2d^2}=boxed{frac{E}{2}}
end{align*}$$

Result

4 of 4

$text{B}.$

Exercise 4

Step 1

1 of 4

The magnitude of the electric field due to a point charge $q$ at a distance $r$ is given by:
$$begin{align*}
E=kfrac{vert q vert}{r^2}
end{align*}$$
where $k=8.99times 10^9;text{N}cdottext{m}^2/text{C}^2$, and $r$ is the distance of the field from the point charge $q$.

Step 2

2 of 4

For this problem, we treat the hollow metal sphere as a point charge with $q=10;mutext{C}$ located at $x=2;text{m}$. Since the electric field is located at $x=8;text{m}$, the distance between the field and the point charge would be $r=6;text{m}$. Substituting these known values, we get:
$$begin{align*}
E &=(8.99times 10^9;text{N}cdottext{m}^2/text{C}^2)frac{vert 10times10^{-6}text{C} vert}{(6;text{m})^2}=2497;frac{text{N}}{text{C}}approx2500;frac{text{N}}{text{C}}
end{align*}$$

Step 3

3 of 4

The direction of an electric field is toward a negative charge and away from a positive charge. Since the point charge $q$ is a positive charge, the electric field will point along the positive direction of $x$.
$$begin{align*}
vec{E}=boxed{2500;frac{text{N}}{text{C}};hat{x}}
end{align*}$$

Result

4 of 4

$text{A.}$

Exercise 5

Step 1

1 of 3

The electric potential energy for point charges $q$ and $q_0$ is given by:
$$begin{align*}
PE=kfrac{q_0 q}{r}
end{align*}$$
where $k=8.99times 10^9;text{N}cdottext{m}^2/text{C}^2$, and $r$ is the distance between the two point charges.

Step 2

2 of 3

We treat the metal hollow sphere as a point charge $q=10;mutext{C}$ located at $x=2;text{m}$. Additionally, we are given a proton, whose charge is $q_0=e=1.6times10^{-19};text{C}$, located at $x=8;text{m}$. Therefore, the distance between these point charges is $x=6;text{m}$. Substituting these values to the equation, we get:
$$begin{align*}
PE&=(8.99times 10^9;text{N}cdottext{m}^2/text{C}^2)frac{(1.6times10^{-19};text{C}) (10times10^{-6};text{C})}{6;text{m}}\&=boxed{2.4times10^{-15};text{J}}
end{align*}$$

Result

3 of 3

$text{B.}$

Exercise 6

Step 1

1 of 5

The magnitude of a charge $Q$ stored on a capacitor is given by:
$$begin{align*}
Q=CV
end{align*}$$
where $C$ is the capacitance and $V$ is the electric potential difference.

Step 2

2 of 5

If we double the potential difference to $2V$, the new charge stored $Q’$ would be:
$$begin{align*}
Q’=C(2V)=2(CV)=boxed{2Q}
end{align*}$$

Step 3

3 of 5

Additionally, the electrical energy is given by:
$$begin{align*}
PE=frac{1}{2}QV
end{align*}$$

Step 4

4 of 5

The new energy $PE’$ would be in terms of $Q’$ when we double the potential difference to $2V$.
$$begin{align*}
PE’=frac{1}{2}Q'(2V)=frac{1}{2}(2Q)(2V)=frac{1}{2}4QV=boxed{4PE}
end{align*}$$

Result

5 of 5

$text{C.}$

Exercise 7

Step 1

1 of 1

Assuming that the two plates are charged differently, the path of a stream of electrons as they enter the uniform electric field between the two charged parallel plates will be straight and **as they move through the field, they will be have a curved path with a direction that follows the direction of the electric field which is going to the positive charged plate**. As it exits the field, it will continue to trace the path of the direction of its last trajectory.

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