In the figure, the four particles form a square of edge length a =

5.00 cm and have charges q1 = +15.0 nC, q2 = -30.0 nC, q3 = +30.0

nC and q4 = -15.0 nC. What net electric field do the particles

produce at the square’s center? ____N/Ci + ___N/Cj

General guidance

[Section: Concepts and reason]

The question is based upon the electric field due to a dipole at a point on the equatorial line.

Initially, draw the labeled diagram of the electric field due to dipole. Find the expressions for the angle made by the field with the vertical and the distance of the charge from the field point.

Then, find the electric field due to one of the dipole at the center of the square. For this write the expression of the electric field due to one of the charge substitute the values of the distance and the angle in the expression. After that take, the vector sum of the electric field due to charges.

Repeat the same with another dipole and take the vector sum of the field due to two dipoles. After that plug in the values in the derived expression of electric field and solve.

Fundamentals

The expression of the electric field is,

Here, E is the electric field, k is the coulomb’s constant, Q is the charge, and R is the distance.

Step-by-step

Step 1 of 2

Consider an electric dipole of charges and placed at distance apart. The electric field at a point on the perpendicular bisector at distance from the center of the dipole is shown in the following diagram.

Here, is the angle made by the electric field with the vertical, is the length of the rod and is the distance between the center of the dipole to the field point.

and are the electric field at the field point due to positive charge and negative charge respectively.

From the figure, the distance of the charge from the field point is,

The angle is given below:

Substitute for as follows:

Let the horizontal direction be taken as axis and vertical direction is component.

From the given diagram, due to symmetry the horizontal component of electric field cancels out each other and vertical component of electric fields add up. Thus, the net electric field due to charges and acts in vertically downward direction.

On the same line, the net electric field due to charges and acts in vertically upward direction.

The component of electric field due to charge is,

Substitutefor and for as follows:

Again substitute for as follows:

So, the net electric field at the center of the square due to charges and is given below:

On the same line, the electric field at the center of the square due to charges and is given as follows:

Hence, the expression for the net electric field at the center of the square is,

The direction of the electric field due to positive charge is directed radially outward from the positive charge and radially inwards for a point negative charge.

Plug in the values in the expression of electric field derived in the last step.

Step 2 of 2

The net electric field at the center of the square is given as follows:

Substitute for and ; for and ; for and for as follows:

Hence, the electric field at the center of the square is .

The electric field at the center of the square is .

The electric field at any point due to the dipole is the vector sum of electric field along and direction. But here due to symmetry, the horizontal component of the electric field cancels each other.

So, the net electric field at the center of the square is equal to the sum of vertical components of electric field.

Answer

The electric field at the center of the square is .

Answer only

The electric field at the center of the square is .

É sin N E sin Ē, E, cos Ecoso Figure 1. showing the electric field at field point.

sin

sin =-

E, = kg sino

”

*)

(5 + b) ਬਹੁ (E) *(E)ਣਾ . “g + “g =”3

0% b) ਖ਼C) (0 +() = += “g

(*6+b+*6–16–) ¥z, (1)(+6+b)xen +()(b + b)xen **1+ “I = 7

(b+ b + b = b-) xyz =

15.0 nC

30.0 nC

5.00 cm

8.987×10°N-m²/C

È = V2k(-9 –94 +92 +93); V2(-15.0-15.0+30.0+30.0)(1×10°C/nC)(8.987×10° N·m²/C) (5.00 cm)’ (0.01 m/cm) =1.53×10 N/C )

ON/C î +1.53×10′ N/C į

ON/C î +1.53×10′ N/C į

ON/C î +1.53×10′ N/C į

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