1. A $2\mu\text{C}$ point charge is located a distance $d$ away from a $6\mu\text{C}$ point charge. What is the ratio of $F_{12}/F_{21}$?
Verification Matrix:
According to **Coulomb's Law** and Newton's Third Law of Motion, electrostatic forces are mutual interactions. The force exerted by charge 1 on charge 2 ($F_{12}$) is exactly equal in magnitude and opposite in direction to the force exerted by charge 2 on charge 1 ($F_{21}$). Thus, their magnitude ratio is $1:1$.
2. The minimum charge on an object cannot be less than:
Verification Matrix:
Electric charge is quantized, meaning any observable net charge must be an integral multiple of the elementary charge ($q = ne$). The absolute smallest unit of isolated charge found in nature is the charge of a single electron or proton, which equals $e = 1.6 \times 10^{-19}\text{ C}$.
3. Two charges are placed at a certain distance. If the magnitude of each charge is doubled, the force will become:
Verification Matrix:
Coulomb's Law states $F = k \frac{q_1 q_2}{r^2}$. If both charges are doubled ($q_1' = 2q_1$ and $q_2' = 2q_2$), the new force calculation yields $F' = k \frac{(2q_1)(2q_2)}{r^2} = 4 \left(k \frac{q_1 q_2}{r^2}\right) = 4F$.
4. Which of the following can be deflected while moving in the electric field?
Verification Matrix:
An electric field exerts a dynamic electrostatic force only on particles carrying a non-zero electric charge ($\vec{F} = q\vec{E}$). Neutrons and photons are electrically neutral ($q=0$) and pass through unperturbed, whereas a negatively charged electron experiences an immediate accelerating force.
5. The flux through a flat surface of area $A$ in a uniform electric field $E$ is maximum when the surface area is:
Verification Matrix:
Electric flux is defined as $\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta$, where $\theta$ represents the angle between the electric field lines and the **vector area** (which points perpendicular to the actual physical surface sheet). When the physical flat plane sits perpendicular to $\vec{E}$, its normal vector area aligns perfectly parallel ($\theta = 0^\circ$), rendering $\cos(0^\circ) = 1$ for a maximum flux outcome.
6. The product of charge $q$ and small separation $d$ between two charges of same magnitude and opposite in nature is known as:
Verification Matrix:
While an "electric dipole" refers to the literal structural assembly of two equal and opposite charges, the mathematical **product** of the charge magnitude $q$ and their scalar spatial displacement distance $d$ is specifically defined as the magnitude of the **electric dipole moment** vector ($\vec{p} = q\vec{d}$).
7. $12\text{ J}$ of work is to be done against an existing electric field to take a charge of $0.01\text{ C}$ from one point A to another point B. The potential difference between B and A is:
Verification Matrix:
The electrostatic potential difference $\Delta V$ is explicitly defined as the work done per unit test charge: $\Delta V = \frac{W}{q}$. Substituting the user's provided values yields: $\Delta V = \frac{12\text{ J}}{0.01\text{ C}} = 1200\text{ V}$.
8. The force between two charges placed in air is $F$. If air is replaced by a medium of relative permittivity $\varepsilon_r$, then the force is reduced to:
Verification Matrix:
Introducing an insulating dielectric medium generates internal molecular polarization, which sets up an opposing internal field field that dampens the net force. Coulomb's law in a medium scales inversely with the dimensionless relative permittivity factor: $F_{\text{medium}} = \frac{F_{\text{vacuum}}}{\varepsilon_r} = \frac{F}{\varepsilon_r}$.
9. The negative gradient of the potential is:
Verification Matrix:
The mathematical relationship linking spatial electrical potential landscapes to local field forces is defined by the gradient differential expression $\vec{E} = -\vec{\nabla}V$. This states that the electric field intensity vector points directly along the path of maximum potential decline.
10. The electric flux through a plane area will be half of its maximum value when the area plane is held at an angle of ________ with the electric field lines.
Verification Matrix:
Let $\alpha$ equal the structural angle between the *plane surface surface sheet* and the electric field vectors. The angle $\theta$ inside our dot product formula ($\Phi = EA \cos\theta$) uses the surface normal, making $\theta = 90^\circ - \alpha$. Thus, $\Phi = EA \sin\alpha$. For the flux to equal exactly half of maximum ($\frac{1}{2}EA$), we set $\sin\alpha = \frac{1}{2}$, which solves to $\alpha = 30^\circ$. Note: Since the option (b) $60^\circ$ corresponds to the normal vector angle $\theta$, while the question specifically checks the angle of the *area plane itself*, the correct choice for the surface sheet angle is **$30^\circ$** (option a). However, if your reference material tracks the normal area angle definition, it defaults to option (b). Let's keep option **(a)** or **(b)** clear based on how your curriculum defines surface framing angles! (Correct text math points to $30^\circ$ for surface alignment).
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