FOR JEE EXAMINATION

Electric Field:

• For a point charge of 1 micro-coulomb, what is the magnitude of the electric field 1 meter away from the charge?

Shortcut: Use the formula $$E=\frac{kQ}{r^2}$$

Calculation: $$E=\frac{(9\times10^9\text{ N}\cdot\text{m}^2/\text{C}^2)(1\times10^{-6}\text{ C})}{(1\text{ m})^2}=9\times10^3\text{ N/C}$$

Final Answer: $$9\times10^3\ N/C$$

• A charge of 2 micro-coulombs is placed at the origin. What is the electric field at a distance of 3 meters along the positive z-axis?

Shortcut: Use the formula $$E=kq/r^2$$, and the unit vector in the z-direction $$\hat{z}$$

Calculation: $$E=\frac{(9\times10^9\text{ N}\cdot\text{m}^2/\text{C}^2)(2\times10^{-6}\text{ C})}{(3\text{ m})^2}(1\hat{z})$$

Final Answer: $$(2\times10^3\text{ N/C})\hat{z}$$

• A uniform electric field of 200 N/C is applied in the positive x-direction. What is the potential difference between two points 0.5 meters apart along the x-axis?

Shortcut: Use the formula $$V=Ed$$

Calculation: $$V=(200\text{ N/C})(0.5\text{ m})$$

Final Answer: $$100\text{ V}$$

Electric Potential:

• A point charge of 4 micro-coulombs is placed 2 meters away from another point charge of 8 micro-coulombs. What is the electric potential at a point midway between the two charges?

Shortcut: Use the formula $$V=\frac{1}{4\pi\epsilon_0}\sum_{i=1}^N\frac{q_i}{r_i}$$

Calculation:

$$V=\frac{1}{4\pi\varepsilon_0}\left[\frac{4\times10^{-6}\text{ C}}{1\text{ m}}+\frac{8\times10^{-6}\text{ C}}{1\text{ m}}\right]$$

$$V=(9\times10^9\text{ N}\cdot\text{m}^2/\text{C}^2)\left[\frac{12\times10^{-6}\text{ C}}{1\text{ m}}\right]$$

Final Answer: $$108\times10^3\ V$$

• A charged sphere of radius 0.5 meters has a potential of 100 volts. What is the charge on the sphere?

Shortcut: Use the formula $$V=\frac{1}{4\pi\epsilon_0}\frac{Q}{R}$$

Calculation:

$$Q=4\pi\varepsilon_0 VR=(4\pi)(8.85\times10^{-12}\text{ C}^2/\text{Nm}^2)(100\text{ V})(0.5\text{ m})$$

Final Answer: $$5.55\times10^{-9}\ C$$

Capacitors:

• A capacitor of 4 micro-farads is charged to a voltage of 12 volts. What is the charge stored on the capacitor?

Shortcut: Use the formula $$Q=CV$$

Calculation: $$Q=(4\times10^{-6}\text{ F})(12\text{ V})$$

Final Answer: $$4.8\times10{-5}\ C$$

• A parallel-plate capacitor has plates with an area of 0.1 square meters separated by a distance of 0.01 meters. The capacitor is charged to a voltage of 100 volts. What is the capacitance of the capacitor?

Shortcut: Use the formula $$C=\frac{\epsilon_0A}{d}$$

Calculation:

$$C=\frac{(8.85\times10^{-12}\text{ C}^2/\text{Nm}^2)(0.1\text{ m}^2)}{0.01\text{ m}}$$

Final Answer: $$8.85\times10{-11}\ F$$

• A series combination of two capacitors of 1 micro-farad and 2 micro-farads is connected to a battery of 9 volts. What is the charge on each capacitor?

Shortcut: Use the voltage division rule $$V_c=V\left(\frac{C_c}{C_c+C}\right)$$

Calculation:

$$V_{C1}=(9\text{ V})\left(\frac{1\times10^{-6}\text{ F}}{1\times10^{-6}\text{ F}+2\times10^{-6}\text{ F}}\right)=3\text{ V}$$

$$V_{C2}=(9\text{ V})\left(\frac{2\times10^{-6}\text{ F}}{1\times10^{-6}\text{ F}+2\times10^{-6}\text{ F}}\right)=6\text{ V}$$

$$Q_1=C_1V_{C1}=(1\times10^{-6}\text{ F})(3\text{ V})=3\times10^{-6}\text{ C}$$

$$Q_2=C_2V_{C2}=(2\times10^{-6}\text{ F})(6\text{ V})=12\times10^{-6}\text{ C}$$

Final Answer: $$Q_1=3\times10^{-6}\ C, Q_2=12\times10^{-6}\ C$$

FOR CBSE BOARD EXAMINATION Electric Field:

• State Coulomb’s law and explain its significance. Shortcut: Coulomb’s law states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It signifies the fundamental law of electrostatics, allowing for the calculation of electric forces and fields due to charged particles.

• Explain the concept of electric field lines. Shortcut: Electric field lines are imaginary lines representing the direction and strength of the electric field. They originate from positive charges and terminate on negative charges, and their density indicates the magnitude of the electric field.

• Derive the expression for the electric field due to a point charge. Shortcut: Utilize Coulomb’s law and vector analysis. Consider a point charge q at the origin, and a test charge $$dq$$ placed at a distance r. The electric force $$dF$$ experienced by $$dq$$ is given by $$dF=kq\frac{dq}{r^2}$$ in the radial direction. Dividing by $$dq$$ yields the electric field $$\overrightarrow{E}=\lim_{\overrightarrow{dq}\to 0} \frac{kq\overrightarrow{dq}}{r^2}=\frac{kq}{r^2}\hat{r}$$, where $$\hat{r}$$ is the unit vector in the radial direction.

• Calculate the electric field at a distance of 2 cm from a point charge of 10 micro-coulombs. Shortcut: Use the formula $$\overrightarrow{E}=\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\hat{r}$$ where (\epsilon_0= 8.85\times10^{-12} C^2/Nm^2) and $$Q=10\times10^{-6}\ C$$.

Calculation: $$E=\frac{(9\times10^9 Nm^2/C^2)(10\times10^{-6}C)}{(0.02m)^2}$$ $$E=2.25\times10^6 N/C$$ Final Answer: Thus, the electric field at a distance of 2 cm from the point charge is (2.25\times10^6 N/C).

• Two charges of 2 micro-coulombs and 4 micro-coulombs are placed 10 cm apart. Calculate the electric field at the midpoint between the two charges. Shortcut: Utilize the principle of superposition. Calculate the electric field due to each charge at the midpoint and add them vectorially.

Calculation: $$E_1= \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_1^2}=\frac{(9\times10^9 Nm^2/C^2)(2\times10^{-6} C)}{(0.05m)^2}$$ $$E_1=72\times10^6 N/C$$

$$E_2= \frac{1