Question:
Two charges \(+2\,\mu\text{C}\) and \(-2\,\mu\text{C}\) are placed \(2\,\text{m}\) apart. Electric field at midpoint is:
Two charges \(+2\,\mu\text{C}\) and \(-2\,\mu\text{C}\) are placed \(2\,\text{m}\) apart. Electric field at midpoint is:
Show Hint
Be careful with signs!
- For opposite charges (an electric dipole), the electric fields reinforce each other at the midpoint (they add up).
- For like charges (e.g., both positive), the fields at the midpoint oppose and cancel each other out to zero.
- For opposite charges (an electric dipole), the electric fields reinforce each other at the midpoint (they add up).
- For like charges (e.g., both positive), the fields at the midpoint oppose and cancel each other out to zero.
Updated On: May 21, 2026
- $0$
- $9 \times 10^3\text{ N/C}$
- $18 \times 10^3\text{ N/C}$
- $36 \times 10^3\text{ N/C}$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks to calculate the net electric field at the exact midpoint of a line segment connecting two equal and opposite point charges.
Step 2: Key Formula or Approach:
The electric field ($E$) produced by a point charge $q$ at a distance $r$ is:
\[ E = \frac{k|q|}{r^2} \] where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$.
The total electric field is the vector sum of the individual fields produced by both charges (Principle of Superposition).
Step 3: Detailed Explanation:
• Let us define the setup:
Charge $q_1 = +2\mu\text{C} = +2 \times 10^{-6}\text{ C}$ is on the left.
Charge $q_2 = -2\mu\text{C} = -2 \times 10^{-6}\text{ C}$ is on the right.
The separation distance is $d = 2\text{ m}$.
• The midpoint lies at a distance of $r = \frac{d}{2} = 1\text{ m}$ from both charges.
• Let us determine the direction of the electric field vectors at this midpoint:
- The positive charge $q_1$ creates an electric field vector $\vec{E}_1$ pointing away from itself (directed to the right).
- The negative charge $q_2$ creates an electric field vector $\vec{E}_2$ pointing towards itself (also directed to the right).
• Since both $\vec{E}_1$ and $\vec{E}_2$ point in the exact same direction (to the right), the net electric field magnitude is the simple scalar sum of their individual magnitudes:
\[ E_{\text{net}} = E_1 + E_2 \]
• Calculating the individual magnitudes:
\[ E_1 = \frac{k \cdot |q_1|}{r^2} = \frac{(9 \times 10^9) \cdot (2 \times 10^{-6})}{1^2} = 18 \times 10^3\text{ N/C} \] \[ E_2 = \frac{k \cdot |q_2|}{r^2} = \frac{(9 \times 10^9) \cdot (2 \times 10^{-6})}{1^2} = 18 \times 10^3\text{ N/C} \]
• Summing these magnitudes:
\[ E_{\text{net}} = 18 \times 10^3 + 18 \times 10^3 = 36 \times 10^3\text{ N/C} \]
• The net field is directed towards the negative charge.
Step 4: Final Answer:
The net electric field at the midpoint is $36 \times 10^3\text{ N/C}$.
The question asks to calculate the net electric field at the exact midpoint of a line segment connecting two equal and opposite point charges.
Step 2: Key Formula or Approach:
The electric field ($E$) produced by a point charge $q$ at a distance $r$ is:
\[ E = \frac{k|q|}{r^2} \] where $k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$.
The total electric field is the vector sum of the individual fields produced by both charges (Principle of Superposition).
Step 3: Detailed Explanation:
• Let us define the setup:
Charge $q_1 = +2\mu\text{C} = +2 \times 10^{-6}\text{ C}$ is on the left.
Charge $q_2 = -2\mu\text{C} = -2 \times 10^{-6}\text{ C}$ is on the right.
The separation distance is $d = 2\text{ m}$.
• The midpoint lies at a distance of $r = \frac{d}{2} = 1\text{ m}$ from both charges.
• Let us determine the direction of the electric field vectors at this midpoint:
- The positive charge $q_1$ creates an electric field vector $\vec{E}_1$ pointing away from itself (directed to the right).
- The negative charge $q_2$ creates an electric field vector $\vec{E}_2$ pointing towards itself (also directed to the right).
• Since both $\vec{E}_1$ and $\vec{E}_2$ point in the exact same direction (to the right), the net electric field magnitude is the simple scalar sum of their individual magnitudes:
\[ E_{\text{net}} = E_1 + E_2 \]
• Calculating the individual magnitudes:
\[ E_1 = \frac{k \cdot |q_1|}{r^2} = \frac{(9 \times 10^9) \cdot (2 \times 10^{-6})}{1^2} = 18 \times 10^3\text{ N/C} \] \[ E_2 = \frac{k \cdot |q_2|}{r^2} = \frac{(9 \times 10^9) \cdot (2 \times 10^{-6})}{1^2} = 18 \times 10^3\text{ N/C} \]
• Summing these magnitudes:
\[ E_{\text{net}} = 18 \times 10^3 + 18 \times 10^3 = 36 \times 10^3\text{ N/C} \]
• The net field is directed towards the negative charge.
Step 4: Final Answer:
The net electric field at the midpoint is $36 \times 10^3\text{ N/C}$.
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Electrostatics Questions
- The electric field intensity (\(E\)) at a distance of 3 m from a uniform long straight wire of linear charge density 0.2 \(\mu C m^{-1}\) is:
- The capacitance of an isolated sphere of radius \( r_1 \) is increased by 5 times when enclosed by an earthed concentric sphere of radius \( r_2 \). The ratio \( \frac{r_1}{r_2} \) is:
- Two spheres A & B of radii 4 cm & 6 cm are given charges of 80 µC & 40 µC respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is:
- The angle between the electric dipole moment of a dipole and the electric field strength due to it on the equatorial line is:
Two condensers \( C_1 \) & \( C_2 \) in a circuit are joined as shown in the figure. The potential of point A is \( V_1 \) and that of point B is \( V_2 \). The potential at point D will be:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions