What does $q_1+q_2=0$ represent in electrostatics?
Answer
$q_1+q_2=0 \Rightarrow q_1=-q_2 i$ e. $q_1$ and $q_2$ are equal in result and opposite in nature. In electronics it actually refers to charge conservation.
$q_1$ and $q_2$ are point charges such that $q_1 \times q_2>9$. What is the nature of the force between the two charges?
Answer
Here, $q_1 q_2>0$. This means that $q_1>0$ and $q_2>0$ or $q_1<0, q_2<0$ when both $q_1$ and $q_2$ are positive or negative. Thus, the foree between $q_1$ and $q_2$ will be repulsive.
Can a metal sphere of radius one meter be given a charge of 1 coulomb?
Answer
No, because $\begin{aligned} E= \frac{k q}{R^2}=\frac{9 \times 10^9 \times 1}{1^2} \\ E=9 \times 10^9 \frac{\text { Volt }}{\text { meter }}\end{aligned}$ will occur on the surface of the sphere. If the electric field in the air exceods $3 \times 10^6$ Volt/ meter, the air will get ionized due to which the charge of the sphere will get destroyed in the air.
How does the electric field due to point charge and line charge change with distance?
Answer
Due to point charge $E =\frac{1}{4 \pi \in_0}\left(\frac{q}{r^2}\right)$ $E \propto \frac{1}{r^2}$ And due to linear charge $E =\frac{\lambda}{2 \pi \in_0 r} \Rightarrow E \propto \frac{1}{r}$
Definition of electric field intensity at a point.
Answer
The electric force acting on a unit test positive charge at any point in the electric field is called the intensity of the electric field at that point. $E =\lim _{q_0 \rightarrow 0}\left(\frac{F}{q_0}\right)$
The force of attraction or repulsion (F) between two point charges is proportional to the product of the magnitude of the charges $q_1$ and $q_2$ and inversely proportional to the square of the distance between the charges $\left(1 / r^2\right)$. That is $F \propto \frac{q_1 q_2}{r^2}$ or $F = K \frac{q_1 q_2}{r^2}$ or $F =\frac{1}{4 \pi \in_o} \frac{q_1 q_2}{r^2}$ Here, $K \left(=\frac{1}{4 \pi \in_{ o }}\right)=9 \times 10^9 Nm ^2 / C ^2$.
If two electric charges $q_1$ and $q_2$ are such that (i) $q_1 q_2>0$, (ii) $q_1 q_2<0$, then what will be the nature of the electric force between them?
Answer
(i) Repulsive (since in this case both the chargers will be of similar nature).
(ii) Attractive (since in this case both the charges will be of opposite nature).