# Electric Field and Potential: Problem 1.08, 1.09 & 1.10 with detailed Solutions

**1.08) A point charge +50 μC is fixed in the xy plane at the point of position vector rᤱ = (2i + 3j) m. What is the intensity of electric field at the point of position vector r = (8i - 5j) m?**

**A) 1200 V/m**

**B) 0.04 V/m**

**C) 900 V/m**

**D) 4500 V/m**

Let the point charge +50 μC is placed at the point A and we have to find the strength of electric field at the point B. Then, as per the statement of the problem, the point A has postion vector (2

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1.09) Four point charges are arranged at the corners of a square ABCD, as shown in the figure.

**i**+ 3**j**) m and that of the point B is (8**i**- 5**j**) m. Or in terms of Cartesian coordinates, these points are A(2, 3) and B(8, -5). Distance between the points A and B can be obtained with the aid of distance formula.
Now, the strength of electric field at a distance of d (= 10 m) from the point charge q (= +50 μC) can be calculated as:

Hence, the intensity of electric field at point B is

**4500 V/m**.###
1.09) Four point charges are arranged at the corners of a square ABCD, as shown in the figure.

The force experienced by a positive charge located at the centre of the square is :
A) zero

B) along diagonal AC

C) along diagonal BD

D) perpendicular to the side AB

Below given diagram shows the forces acting on the point charge located at the centre of the square. Length of the arrow represents magnitude.
The direction of resultant force (R) is perpendicular to the side AB.

1.10) A particle of mass m and charge q is kept in a sinusoidally varying electric field E' = Esinωt. It will undergo a simple harmonic motion of amplitude

A) qE²/mω²

B) qE/mω²

C) (qE/mω²)⁰·⁵

D) qE/mω

The force experienced by a positive charge located at the centre of the square is :

- A) zero

- B) along diagonal AC

- C) along diagonal BD

- D) perpendicular to the side AB

1.10) A particle of mass m and charge q is kept in a sinusoidally varying electric field E' = Esinωt. It will undergo a simple harmonic motion of amplitude

- A) qE²/mω²

- B) qE/mω²

- C) (qE/mω²)⁰·⁵

- D) qE/mω

The electric field at a general instant t = t is E'(t) = Esinωt. So, the force experienced by the point charge q as a function of time t is F(t) = qE(t) = qEsinωt. If 'a' be the resulting acceleration of the particle, then by Newton's second law of motion, a(t) = F(t)/m = (qE/m)sinωt, where m is the mass. From this expression we observe that the acceleration holds a sinusoidal relationship with time which means the particle performs simple harmonic motion (SHM).

From the obtained expression of acceleration, the maximum magnitude of acceleration aᤱ = qE/m. Also, in SHM the same is also given by aᤱ = Aω²,

*A*being the amplitude of oscillation. On equating the two, we get the amplitude**A = qE/mω²**.