A vertical electric field of magnitude 4.9 × 105 N/C just prevents a water droplet of a mass of 0.1 g from falling. The value charge on the droplet will be(Given g = 9.8 m/s2)
1.6 × 10–9 C
2.0 × 10–9 C
3.2 × 10–9 C
0.5 × 10–9 C
The Correct Option is B
Solution and Explanation
Since the droplet is at rest
⇒ Net force = 0
⇒ mg = qE
⇒ q=\(\frac{mg}{E}\)=2×10-9 C
The correct option is (B) : 2.0 × 10–9 C
Top Questions on Electric Field
- The electric field (\( \vec{E} \)) and electric potential (\( V \)) at a point inside a charged hollow metallic sphere are respectively:
- CBSE CLASS XII - 2025
- Physics
- Electric Field
- Consider a circular loop that is uniformly charged and has a radius $ \sqrt{2} $. Find the position along the positive $ z $-axis of the cartesian coordinate system where the electric field is maximum if the ring was assumed to be placed in the $ xy $-plane at the origin:
- JEE Main - 2025
- Physics
- Electric Field
Two large plane parallel conducting plates are kept 10 cm apart as shown in figure. The potential difference between them is $ V $. The potential difference between the points A and B (shown in the figure) is:
- JEE Main - 2025
- Physics
- Electric Field
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason
(R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions.
In the light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2025
- Physics
- Electric Field
A metallic ring is uniformly charged as shown in the figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to O is ‘E’ magnitude. What would be the magnitude of electric field at ‘O’ due to arc ABC?
- JEE Main - 2025
- Physics
- Electric Field
Questions Asked in JEE Main exam
Match the LIST-I with LIST-II for an isothermal process of an ideal gas system.
Choose the correct answer from the options given below:
- JEE Main - 2026
- Thermodynamics and Work Done
- A bag contains 10 balls out of which \( k \) are red and \( (10-k) \) are black, where \( 0 \le k \le 10 \). If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:
- JEE Main - 2026
- Probability
Which one of the following graphs accurately represents the plot of partial pressure of CS₂ vs its mole fraction in a mixture of acetone and CS₂ at constant temperature?
- JEE Main - 2026
- Organic Chemistry
- For two identical cells each having emf \(E\) and internal resistance \(r\), the current through an external resistor of \(6\,\Omega\) is the same when the cells are connected in series as well as in parallel. The value of the internal resistance \(r\) is ________ \(\Omega\).
- JEE Main - 2026
- Current electricity
- For three unit vectors \( \vec a, \vec b, \vec c \) satisfying \[ |\vec a-\vec b|^2 + |\vec b-\vec c|^2 + |\vec c-\vec a|^2 = 9 \] and \[ |2\vec a + k\vec b + k\vec c| = 3, \] the positive value of \( k \) is:
- JEE Main - 2026
- Vector Algebra
Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).