Using Coulomb's law, $F=kq_1q_2/r^2$. Here $q_1=2\times10^{-7}\,\text{C}$, $q_2=3\times10^{-7}\,\text{C}$, $r=0.30\,\text{m}$ and $k=9\times10^9\,\text{N m}^2\text{C}^{-2}$. Thus $F=9\times10^9(2\times10^{-7})(3\times10^{-7})/(0.30)^2=6.0\times10^{-3}\,\text{N}$. Since both charges are positive, the force is repulsive.
$6.0\times10^{-3}\,\text{N}$, repulsive.
From $F=k|q_1q_2|/r^2$, $r=\sqrt{k|q_1q_2|/F}$. With $q_1=0.4\times10^{-6}\,\text{C}$ and $q_2=0.8\times10^{-6}\,\text{C}$, $r=\sqrt{(9\times10^9)(0.4\times10^{-6})(0.8\times10^{-6})/0.2}=0.12\,\text{m}$. By Newton's third law, the second sphere experiences a force equal in magnitude and opposite in direction, so it is $0.2\,\text{N}$ and attractive.
(a) $0.12\,\text{m}$; (b) $0.2\,\text{N}$, opposite in direction to the force on the first sphere.
Both $ke^2/r^2$ and $Gm_em_p/r^2$ are forces, so their ratio $ke^2/(Gm_em_p)$ has no dimensions. Using $k=8.99\times10^9$, $e=1.60\times10^{-19}\,\text{C}$, $G=6.67\times10^{-11}$, $m_e=9.11\times10^{-31}\,\text{kg}$ and $m_p=1.67\times10^{-27}\,\text{kg}$ gives $ke^2/(Gm_em_p)\approx2.3\times10^{39}$.
The ratio is dimensionless and is about $2.3\times10^{39}$. It signifies that the electrostatic force between an electron and a proton is about $10^{39}$ times stronger than their gravitational attraction.
Quantisation means that charge cannot take arbitrary values; it occurs as $q=ne$, where $n$ is an integer and $e=1.6\times10^{-19}\,\text{C}$. In ordinary macroscopic situations, the number of elementary charges involved is enormous. The step size $e$ is then too small compared with the total charge to be detected, so charge can be treated as continuous.
(a) A body's charge is an integral multiple of the elementary charge: $q=ne$. (b) For macroscopic charges, $n$ is extremely large, so the charge appears continuous.
When glass is rubbed with silk, electrons are transferred from one body to the other. The glass and silk acquire equal and opposite charges, so the algebraic sum of their charges remains zero if it was zero initially. Thus the appearance of charge on both bodies is consistent with conservation of total charge.
Rubbing transfers charge from one body to the other; it does not create net charge.
Opposite corners A and C carry equal charges, so their forces on the central charge are equal in magnitude and opposite in direction. Opposite corners B and D also carry equal charges, so their forces on the central charge cancel. Therefore the vector sum of all four forces is zero.
Zero.
The tangent to a field line at any point represents the direction of the electric field there. Since the electric field has a definite direction at every ordinary point in space, a field line must be continuous except where it begins or ends on charge. Two field lines cannot cross because a crossing would assign two different tangents, and therefore two different field directions, at the same point.
(a) A field line gives the direction of electric field at each point, so it cannot abruptly stop or break in a charge-free region. (b) If two field lines crossed, the electric field at the crossing point would have two directions, which is impossible.
At the midpoint, each charge is $0.10\,\text{m}$ away. The fields due to $+3\,\mu\text{C}$ and $-3\,\mu\text{C}$ are both directed from A to B, so $E=2kq/r^2=2(9\times10^9)(3\times10^{-6})/(0.10)^2=5.4\times10^6\,\text{N C}^{-1}$. The force on a charge $q_0=-1.5\times10^{-9}\,\text{C}$ is $F=q_0E$, so its magnitude is $1.5\times10^{-9}\times5.4\times10^6=8.1\times10^{-3}\,\text{N}$ and its direction is opposite to $E$.
(a) $5.4\times10^6\,\text{N C}^{-1}$ along AB from the positive charge to the negative charge. (b) $8.1\times10^{-3}\,\text{N}$, opposite to the field direction.
The two charges are equal and opposite, so total charge is zero. The separation from negative charge at B to positive charge at A is $0.30\,\text{m}$ in the negative z-direction. Therefore $\mathbf{p}=q\mathbf{d}=2.5\times10^{-7}\times0.30(-\hat{k})=-7.5\times10^{-8}\hat{k}\,\text{C m}$.
Total charge is zero; dipole moment is $-7.5\times10^{-8}\,\hat{k}\,\text{C m}$.
The torque magnitude on a dipole is $\tau=pE\sin\theta$. Thus $\tau=(4\times10^{-9})(5\times10^4)\sin30^\circ=1.0\times10^{-4}\,\text{N m}$.
$1.0\times10^{-4}\,\text{N m}$.
The number of excess electrons is $n=Q/e=3\times10^{-7}/(1.6\times10^{-19})=1.9\times10^{12}$. Since polythene becomes negatively charged, electrons move from wool to polythene. The transferred mass is $nm_e=(1.9\times10^{12})(9.11\times10^{-31})\approx1.7\times10^{-18}\,\text{kg}$, which is negligible.
(a) About $1.9\times10^{12}$ electrons are transferred from wool to polythene. (b) Yes, but the mass transferred is only about $1.7\times10^{-18}\,\text{kg}$.
For part (a), $F=kq^2/r^2=9\times10^9(6.5\times10^{-7})^2/(0.50)^2=1.52\times10^{-2}\,\text{N}$. In part (b), each charge is doubled, which multiplies $q_1q_2$ by 4, and the separation is halved, which multiplies $1/r^2$ by 4. Hence the force is $16F=16(1.52\times10^{-2})=2.43\times10^{-1}\,\text{N}$.
(a) $1.52\times10^{-2}\,\text{N}$; (b) $2.43\times10^{-1}\,\text{N}$.
The square has area $A=(0.10)^2=0.010\,\text{m}^2$. Flux is $\Phi=EA\cos\theta$. For a plane parallel to the yz-plane, the normal is along the x-axis, so $\theta=0^\circ$ and $\Phi=3\times10^3\times0.010=30\,\text{N m}^2\text{C}^{-1}$. If the normal makes $60^\circ$ with the x-axis, $\Phi=EA\cos60^\circ=15\,\text{N m}^2\text{C}^{-1}$.
(a) $30\,\text{N m}^2\text{C}^{-1}$; (b) $15\,\text{N m}^2\text{C}^{-1}$.
In a uniform electric field, as much flux enters the cube through one face as leaves through the opposite face. Since there is no net charge enclosed by the cube, Gauss's law also gives net flux $q_{\text{enc}}/\epsilon_0=0$.
Zero.
Gauss's law gives $q_{\text{enc}}=\epsilon_0\Phi=(8.854\times10^{-12})(8.0\times10^3)=7.1\times10^{-8}\,\text{C}$. If the net flux were zero, the algebraic sum of charges inside would be zero, but there could still be equal positive and negative charges inside.
(a) $7.1\times10^{-8}\,\text{C}$. (b) No; zero net flux only means zero net enclosed charge.
The charge is at the centre of a cube of side $10\,\text{cm}$ if the given square is one face. By symmetry, the total flux $q/\epsilon_0$ is equally shared by six faces. Hence $\Phi=q/(6\epsilon_0)=10\times10^{-6}/[6(8.854\times10^{-12})]=1.9\times10^5\,\text{N m}^2\text{C}^{-1}$.
$1.9\times10^5\,\text{N m}^2\text{C}^{-1}$.
The net flux through any closed Gaussian surface enclosing charge $q$ is $q/\epsilon_0$, independent of the size or shape of the surface. Thus $\Phi=2.0\times10^{-6}/(8.854\times10^{-12})=2.26\times10^5\,\text{N m}^2\text{C}^{-1}$.
$2.3\times10^5\,\text{N m}^2\text{C}^{-1}$.
For a closed surface centred on the point charge, flux depends only on enclosed charge, not on radius. Therefore doubling the radius leaves the flux $-1.0\times10^3\,\text{N m}^2\text{C}^{-1}$. From Gauss's law, $q=\epsilon_0\Phi=(8.854\times10^{-12})(-1.0\times10^3)=-8.9\times10^{-9}\,\text{C}$.
(a) $-1.0\times10^3\,\text{N m}^2\text{C}^{-1}$; (b) $-8.9\times10^{-9}\,\text{C}$.
Outside a charged conducting sphere, the field is as if all charge were concentrated at the centre: $E=k|q|/r^2$. Thus $|q|=Er^2/k=(1.5\times10^3)(0.20)^2/(9\times10^9)=6.7\times10^{-9}\,\text{C}$. Since the field points inward, the charge is negative.
$-6.7\times10^{-9}\,\text{C}$.
The radius is $R=1.2\,\text{m}$ and surface area is $4\pi R^2=4\pi(1.2)^2=18.1\,\text{m}^2$. Charge $q=\sigma A=(80.0\times10^{-6})(18.1)=1.45\times10^{-3}\,\text{C}$. The total outward flux is $q/\epsilon_0=1.45\times10^{-3}/(8.854\times10^{-12})=1.64\times10^8\,\text{N m}^2\text{C}^{-1}$.
(a) $1.45\times10^{-3}\,\text{C}$; (b) $1.64\times10^8\,\text{N m}^2\text{C}^{-1}$.
For an infinite line charge, $E=\lambda/(2\pi\epsilon_0 r)$. Hence $\lambda=E(2\pi\epsilon_0 r)=(9\times10^4)[2\pi(8.854\times10^{-12})(0.020)]=1.0\times10^{-7}\,\text{C m}^{-1}$.
$1.0\times10^{-7}\,\text{C m}^{-1}$.
For two large oppositely charged parallel plates with equal surface charge densities on the inner faces, the fields outside cancel. Therefore the field in each outer region is zero. Between the plates, the two fields add and $E=\sigma/\epsilon_0=(17.0\times10^{-22})/(8.854\times10^{-12})=1.92\times10^{-10}\,\text{N C}^{-1}$, directed from the positive plate toward the negative plate.
(a) $0$; (b) $0$; (c) $1.92\times10^{-10}\,\text{N C}^{-1}$, directed from the positive plate to the negative plate.