Electric field
Electric field due to a Positive Charge Electric field due to a Negative Charge
Note:
1. It is important to note that with every charged particle, there is an electric field associated which extends up to infinity.2. No charged particle experiences force due to its own electric field.
Suppose the point charge +Q is located at A, where OA = r1. To calculate the electric field intensity (E) at B, where OB = r2.
Electric field due to Point Charge
Therefore,
Electric field, E=
The electric field intensity at any point due to a system or group of charges is equal to the vector sum of electric field intensities due to individual charges at the same point.
Now, we would do the vector sum of electric field intensities:
E= E1+E2+E3+......+En
Here, ri is the distance of the point P from the ith charge Qi. ri cap is a unit vector directed from Qi to point P. Let’s say charge Q1, Q2...Qn are placed in a vacuum at positions r₁, r₂,....,rₙ respectively.
The net forces at P are the vector sum of forces due to individual charges, given by,
Positive charge q is distributed uniformly along a line with length 2a, lying along the y-axis between y=-a and y=+a. We are here interested in finding the electric field at point P on the x-axis.
λ=charge per unit length= q/2a
dq=λdy
Electric field due to Line Charge Density
Here it is important to note that the x-axis is the perpendicular bisector of our line segment. So, our point P is at a distance x from the midpoint O of the charged line segment.
The distance r from the segment at height y to the field point P is: r=√x2+y2
So, the magnitude of the field at point P due to the segment of height y is
From the figure given above, we can see the x and y components of this field. Here component dEx is perpendicular to the charged line segment and dEy is parallel to it. On resolving dE into x and y components, we get
dEx=dEcosθ
dEy=−dEsinθ
From the symmetry of the configuration, the component of dE parallel to the line charge distribution is zero. When a positive test charge is placed at P, the upper half of the charge line applies force on it along the downward direction, while the lower half applies a force of equal magnitude in an upward direction. This, the top and lower parts of the segment contribute equally to the total field at P, according to symmetry.
Q1: A uniform electric field of 10 N/C is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is (in degree).
Ans: 45
Sol:
Q2: A point charge of 10 μC is placed at the origin. At what location on the X-axis should a point charge of 40pC be placed so that the net electric field is zero at x = 2 cm on the X-axis ? [JEE Main 2023]
(a) x = 6 cm
(b) x = 4 cm
(c) x = 8 cm
(d) x = -4 cm
Ans: (a)
Sol:
Q3: A thin infinite sheet charge and an infinite line charge of respective charge densities +σ and +λ are placed parallel at 5 m distance from each other. Points P and Q are at 3/π m and 4/π m perpendicular distances from line charge towards sheet charge, respectively Ep and Eq are the magnitudes of resultant electric field intensities at point P and Q respectively. If for 2|σ| = |λ|, then the value of a is _____. [JEE Main 2023]
Ans: 6
Sol: Let's consider the following diagram:
The formula to calculate the electric field due to an infinitely long charged sheet is given by
The formula to calculate the electric field due to an infinitely long line charge is given by
Using equations (1) and (2), the electric field at can be calculated as follows:
Similarly, the electric field at is given by
Divide equation (3) by equation (4) to obtain the required ratio.
Hence, a = 6.
The physical significance of electric field is that we can readily calculate the magnitude and direction of force experienced by any charge q0 placed at a point by knowing the electric field intensity at that point.
Electric field lines are an imaginary tool used to visualize the electric field around a charged object.
Q3: Why have we defined the concept of the electric field? Is it really necessary? If eventually, we are measuring the electrostatic force, why can’t we do it directly using Coulomb’s Force?
Sol: The concept of electric field helps us understand the electrical environment around charged objects. It's not always necessary, but it becomes important when we deal with moving charges and electromagnetic waves. When charges are in motion, there is a delay between the cause and effect of their interactions, which can only be explained through the concept of the electric field. Electric and magnetic fields are considered real and physical entities that can transport energy and have their own set of laws. The concept of the electric field is now central in physics.
Electric field lines have several important properties:
Electric field lines of +2Q charge are twice in number than that of +Q. So, irrespective of the number of lines in each representation, the ratio must be maintained to 2.
The relative density of field lines is inversely proportional to the square of the distance. Mathematically, the number of lines per unit area at distance r > number of lines per unit area at distance 2r > the number of lines per unit area at a distance 3r measured from S.
Relative density of field lines at distance r/Relative density of field lines at distance 2r = 4/1
Similarly, Relative density of field lines at distance 2r/Relative density of field lines at distance 3r = 9/4
Electric field lines are a way of pictorially mapping the electric field around a configuration of charges.
(a) Electric field versus q for a positive point charge kept at the origin. Note that the electric field at positive q is positive because it is in a positive direction. At negative q it is negative because it is in a negative direction.
(b) Electric field versus q for a negative point charge kept at the origin. Note that the electric field at positive q is negative because it is in the negative direction. At negative q it is positive because it is in a positive direction.
Q4: Two charged particles lie along the x-axis as shown in the figure. The particle with charge q2 = +8μC is at x = 6.00 m, and the particle with charge q1 = + 2μC is at the origin. Locate the point where the resultant electric field is zero.
Sol:
Before calculating, let us physically see the location of the point where the electric field can be zero. At points other than the x-axis, say above the x-axis, both the charges will have a component of the electric field in the positive direction. This y component of the electric field does not cancel out. So the net electric field at that point will not be zero. The same statement also holds for points that are not in xy -plane.
On the x-axis also we can see that on the points beyond x = 6 m and points on the negative x-axis, both the electric fields will be in the same direction. So the net electric field cannot be zero. At some point between the two charges, the electric field due to both of them will be in opposite directions. So the electric field will be zero at a point between x = 0 and x = 6m.
Let the x coordinate of neutral point (E = 0) be x,
Therefore,
By solving the above equation for x, we get x = 2m.
Hence, at x = 2 between these two charged particles, there exists a point where net electric field due is zero i.e. neutral point.
1. What is an electric field?
Ans. An electric field refers to the region around an electrically charged object or particle where it exerts a force on other charged objects or particles. It is a vector quantity and is represented by an arrow pointing in the direction of the force that a positive test charge would experience if placed in the field.
2. How is the direction of an electric field determined?
Ans. The direction of an electric field is determined by the direction of the force that a positive test charge would experience if placed in the field. The direction of the electric field lines also follows this direction, pointing away from positive charges and towards negative charges.
3. How is the electric field strength due to a point charge calculated?
Ans. The electric field strength due to a point charge can be calculated using the formula E=kq/r^2, where E is the electric field strength, k is Coulomb's constant (9 x 10^9 Nm^2/C^2), q is the charge of the point charge, and r is the distance between the point charge and the location at which the electric field strength is being calculated.
4. What are electric field lines?
Ans. Electric field lines are visual representations of the electric field around a charged object or particle. They are drawn as continuous lines that start from positive charges and end at negative charges, and the density of the lines indicates the strength of the electric field at different points in space.
5. How is the concept of superposition used to calculate the electric field due to multiple charges?
Ans. The concept of superposition is used to calculate the electric field due to multiple charges by calculating the electric field due to each charge and then summing the vectors to find the net electric field. This is based on the principle that the electric field at any point in space is the vector sum of the electric fields due to all the charges present in that space.
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1. What is the formula for calculating the electric field due to a point charge? |
2. How do you determine the direction of the electric field produced by a point charge? |
3. How is the electric field due to a system of point charges calculated? |
4. What are electric field lines and what do they represent? |
5. What are the properties of electric field lines? |
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