Advertisements
Advertisements
Question
Consider a system of n charges q1, q2, ... qn with position vectors `vecr_1,vecr_2,vecr_3,...... vecr_n`relative to some origin 'O'. Deduce the expression for the net electric field`vec E` at a point P with position vector `vecr_p,`due to this system of charges.
Advertisements
Solution
Let us consider a system of n charges q1, q2, ... qn with position vectors r1, r2, r3, ...rn relative to origin O.

Let `vecF_i` be the force due to ith charge qi on q0.
Then,
`vecF_i = 1/(4πε_0) (q_1q_0)/(r_i^2) \ hat r_i`
Here, ri is the distance of the test charge q0 from qi.
The electric field at the observation point P is given by `vecE_i = lim_(q0->0) vecF_i/(q_0) = lim _(q0->0) 1/q_0 (1/(4πε_0) (q_1q_0)/r_1^2 hatr_i)`
`vecE_i = 1/(4πε_0) q_i/(r_1^2)hat r_i ........ (1) `
If `vecE` is the electric field at point P due to the system of charges, then by the principle of superposition of electric fields,
`vecE = vecE_1 +vecE_2 +vecE_3 +....vecE_n = sum_(i=1)^n vecE_i`
Using (1), we get
`vecE = sum_(i=1)^n 1/(4πε_0)q_i/(r_i^2)hat r_i `
`vecE =1/(4πε_0) sum_(i=1)^n q_i/(r_i^2)hat r_i`
APPEARS IN
RELATED QUESTIONS
Show that if we connect the smaller and the outer sphere by a wire, the charge q on the former will always flow to the latter, independent of how large the charge Q is.
If a body is charged by rubbing it, its weight
Consider the situation in the figure. The work done in taking a point charge from P to Ais WA, from P to B is WB and from P to C is WC.

Electric potential decreases uniformly from 120 V to 80 V, as one moves on the x-axis from x = −1 cm to x = +1 cm. The electric field at the origin
(a) must be equal to 20 Vcm−1
(b) may be equal to 20 Vcm−1
(c) may be greater than 20 Vcm−1
(d) may be less than 20 Vcm−1
The electric field in a region is directed outward and is proportional to the distance rfrom the origin. Taking the electric potential at the origin to be zero,
An electric field of 20 NC−1 exists along the x-axis in space. Calculate the potential difference VB − VA where the points A and B are
(a) A = (0, 0); B = (4 m, 2m)
(b) A = (4 m, 2 m); B = (6 m, 5 m)
(c) A = (0, 0); B = (6 m, 5 m)
Do you find any relation between the answers of parts (a), (b) and (c)?
When 1014 electrons are removed from a neutral metal sphere, the charge on the sphere becomes ______.
The Electric field at a point is ______.
- always continuous.
- continuous if there is no charge at that point.
- discontinuous only if there is a negative charge at that point.
- discontinuous if there is a charge at that point.
Five charges, q each are placed at the corners of a regular pentagon of side ‘a’ (Figure).

(a) (i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the corners (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by –q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its corners?
