All Solutions
Page 565: Practice Problems
$$E=frac{F}{q}$$
$$E=frac{2cdot10^{-4}}{5cdot10^{-6}}$$
$$boxed{E=40,,rm N/C}$$
$$
F = q E
$$
Let’s express electric field $E$ from the equation above and solve for it:
$$
begin{align*}
E &= dfrac{F}{q} \
E &= dfrac{ 2 cdot 10^{-4} ~mathrm{N} }{ 5 cdot 10^{-6} ~mathrm{C}}
end{align*}
$$
$$
boxed{E = 40 ~mathrm{dfrac{N}{C}}}
$$
E = 40 ~mathrm{dfrac{N}{C}}
$$
$$E=frac{F}{q}$$
When you put the numbers in you have to put negative sign for charge because it is negative.
$$E=frac{0.06}{-2cdot10^{-8}}$$
$$boxed{E=-3cdot10^{6},,rm N/C}$$
Minus sign means that direction of the field is opposite to the force, because we chose to put direction of the force as positive, hence direction is to the left.
Note that in problem text it’s stated that charge is negative, but we’re told that charge is $2 cdot 10^{-8} ~mathrm{C}$, from which we concluded that charge is
$q = -2 cdot 10^{-8} ~mathrm{C}$, as stated above.
To find the magnitude of the electric field, let’s remind ourselves that force $F$ acting on a charge $q$ in the electric field $E$ is equal to:
$$
F = q E
$$
Notice that since charge $q$ is negative, electric field $E$ must also be negative since we’re given force $F$ and, as we can see, force $F$ is positive. Knowing that electric field $E$ is negative, we conclude that source of the electric field is negatively charged. Let’s express electric field $E$ from the equation above and solve for it:
$$
begin{align*}
E &= dfrac{F}{q} \
E &= dfrac{ 0.06 ~mathrm{N} }{ -2 cdot 10^{-8} ~mathrm{C}}
end{align*}
$$
$$
boxed{ E = – 3 cdot 10^6 ~mathrm{dfrac{N}{C}} }
$$
We are told that direction of the force is to the right and we concluded that this force is repulsive. In order for this force to be repulsive, electric field must be negative, which means that source of the electric field is negatively charged. Direction of the electric field coming from a negatively charged source is always pointing inward to the source, from which we conclude that direction of the electric field is to the left.
E = – 3 cdot 10^6 ~mathrm{dfrac{N}{C}}
$$
$$
text{Direction of electric field is to the left}
$$
$$E=frac{F}{q}rightarrow F=Eq$$
$$F=3cdot10^{-7}cdot27$$
$$boxed{F=8.1cdot10^{-6},,rm N}$$
$$
F = q E
$$
Let’s express electric field $E$ from the equation above and solve for it:
$$
begin{align*}
F &= q E \
F &= 3 cdot 10^{-7} ~mathrm{C} cdot 27 ~mathrm{dfrac{N}{C}}
end{align*}
$$
$$
boxed{ F = 8.1 cdot 10^{-6} ~mathrm{N} }
$$
F = 8.1 cdot 10^{-6} ~mathrm{N}
$$
$$E=frac{F}{q}rightarrow q=frac{F}{E}$$
Force that we put in the equation is equal in magnitude but in opposite in direction to he force of the gravity:
$$F=-F_g$$
$$q=frac{-F_g}{E}$$
$$q=frac{-2.1cdot10^{-3}}{6.5cdot10^{4}}$$
$$boxed{q=-3.23cdot10^{-8},,rm C}$$
$E = 6.5 cdot 10^4 ~mathrm{dfrac{N}{C}}$. If we want to suspend the electric force $F_e$ against force of gravity $F_g$, we need to charge the pith ball in such a way that electric force points upward. Since electric field points downward and we know that force $F$ acting on charge $q$ in an electric field $E$ is equal to:
$$
F_e = q E
$$
we conclude that we need to charge the pith ball negatively, so that electric field and electric force are in the opposite direction.
To calculate charge $q$, we’ll express it from the equation above:
$$
begin{align*}
F_e &= q E \
q &= dfrac{F_e}{E} \
&text{Notice that $F_e = – F_g$} \
q &= dfrac{ – F_g }{E} \
q &= dfrac{ – 2.1 cdot 10^{-3} ~mathrm{N}}{ 6.5 cdot 10^4 ~mathrm{dfrac{N}{C}} }
end{align*}
$$
$$
boxed{ q = -3.23 cdot 10^{-8} ~mathrm{C} }
$$
q = – 3.23 cdot 10^{-8} ~mathrm{C}
$$
You would not measure the same force for the same place with the two test charges. Because force is given with equation:
$$F=Eq$$
we can see that with different charge you get different force.
Yes you would find the same magnitude of the field. Because field magnitude depends only on the position and we did not change the position.
b) Yes
No. The electric force is proportional to the magnitude of the charge.
b)
Yes. The field strength is independent of the magnitude of the charge.
b) Yes.
$$E=kfrac{q}{r^2}$$
where $k$ is Coulomb constant.
$$E=frac{9cdot10^{9}cdot4.2cdot10^{-6}}{1.2^2}$$
$$boxed{E=26250,,rm N/C}$$
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
To calculate the magnitude of the electric field coming from charge
$q = 4.2 cdot 10^{-6} ~mathrm{C}$ at a distance $r = 1.2 ~mathrm{m}$ from it, we will plug in the given data into the equation above:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot4.2 cdot 10^{-6} ~mathrm{C} }{ (1.2 ~mathrm{m})^2 }
end{align*}
$$
$$
boxed{E = 26250 ~mathrm{dfrac{N}{C}}}
$$
E = 26250 ~mathrm{dfrac{N}{C}}
$$
$$E=kfrac{q}{r^2}$$
where $k$ is Coulomb constant.
$$E=frac{9cdot10^9cdot4.2cdot10^{-6}}{2.4^2}$$
$$boxed{E=6562.5,,rm N/C}$$
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
To calculate the magnitude of the electric field coming from charge
$q = 4.2 cdot 10^{-6} ~mathrm{C}$ at a distance $r = 2 cdot 1.2 ~mathrm{m} = 2.4 ~mathrm{m}$ from it, we will plug in the given data into the equation above:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot4.2 cdot 10^{-6} ~mathrm{C} }{ (2.4 ~mathrm{m})^2 }
end{align*}
$$
$$
boxed{E = 6562.5 ~mathrm{dfrac{N}{C}}}
$$
Note that since magnitude of the electric field is inversely proportional to the square of the distance $r$ to the source charge $q$, we expected that magnitude of the electric field will be reduced by a factor of 4 if distance from the source charge was doubled, which is precisely what we calculated.
E = 6562.5 ~mathrm{dfrac{N}{C}}
$$
$$E=kfrac{q}{r^2}$$
where $k$ is Coulomb constant.
$$E=frac{9cdot10^9cdot7.2cdot10^{-6}}{1.6^2}$$
$$boxed{25312.5,,rm N/C}$$
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
To calculate the magnitude of the electric field coming from charge
$q = 7.2 cdot 10^{-6} ~mathrm{C}$ at a distance $r = 1.6 ~mathrm{m}$ from it, we will plug in the given data into the equation above:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
E &= dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 7.2 cdot 10^{-6} ~mathrm{C} }{ (1.6 ~mathrm{m})^2 }
end{align*}
$$
$$
boxed{E = 25312.5 ~mathrm{dfrac{N}{C}}}
$$
Note that magnitude of the electric field doesn’t depend on the direction to in which we measure electric field, which means that when we’re told to calculate the electric field at a point east of the source of the electric field, it gives us an information about direction of the electric field, but not its magnitude. In this case, since source charge is positive, electric field points to the east.
E = 25312.5 ~mathrm{dfrac{N}{C}}
$$
$$
text{Electric field points to the east.}
$$
$$E=kfrac{q}{r^2}rightarrow q=frac{Er^2}{k}$$
where $k$ is Coulomb constant. And because electric field is toward the sphere it means that it has negative sign.
$$q=frac{-450cdot0.25^2}{9cdot10^{9}}$$
$$boxed{q=-3.125cdot10^{-9},,rm C}$$
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
To calculate charge on the sphere for which magnitude of the electric field is $E = 450 ~mathrm{dfrac{N}{C}}$ at a distance $r = 0.25 ~mathrm{m}$ from it, we’ll plug in the given data into the equation above and express charge $q$ from it:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
&tag{multiply by $r^2$} \
Er^2 &= kq\
&tag{divide by $q$}\
q &= dfrac{E r^2}{k} \
q &= dfrac{ 450 ~mathrm{dfrac{N}{C}} cdot (0.25 ~mathrm{m})^2 }{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} } \
q &= 3.125 cdot 10^{-9 } ~mathrm{C}
end{align*}
$$
We’ve calculated the magnitude of the charge on the sphere, but notice that we’re told that electric field points to the sphere, which means that sphere is negatively charged. We thus conclude that charge on the sphere is:
$$
boxed{ q = – 3.125 cdot 10^{-9 } ~mathrm{C} }
$$
q = – 3.125 cdot 10^{-9 } ~mathrm{C}
$$
$$E=kfrac{q}{r^2}rightarrow r=sqrt{frac{kq}{E}}$$
where $k$ is Coulomb constant.
$$r=sqrt{frac{9cdot10^{9}cdot2.4cdot10^{-6}}{360}}$$
$$boxed{r=7.75,,rm m}$$
$$
E = dfrac{kq}{r^2}
$$
where $k = 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}}$ is Coulomb constant.
To calculate distance $r$ where test charge must be placed in order for magnitude of the electric field, coming from charge $q = 2.4 cdot 10^{-6} ~mathrm{C}$ to be $E = 360 ~mathrm{dfrac{N}{C}}$, we’ll use the equation above and express distance $r$ from it:
$$
begin{align*}
E &= dfrac{kq}{r^2} \
&tag{multiply by $r^2$} \
Er^2 &= kq\
&tag{divide by $E$}\
r^2 &= dfrac{kq}{E} \
tag{ find square root of the equation }\
r &= sqrt{ dfrac{kq}{E} } \
r &= sqrt{ dfrac{ 9 cdot 10^9 ~mathrm{dfrac{N m^2}{C^2}} cdot 2.4 cdot 10^{-6} ~mathrm{C} }{ 360 ~mathrm{dfrac{N}{C}} } } \
r &= sqrt{60 ~mathrm{m^2}}
end{align*}
$$
$$
boxed{ r = 7.746 ~mathrm{m} }
$$
r = 7.746 ~mathrm{m}
$$
