Electric Charges and Fields Test

Result

Electric Charges and Fields Test - 1

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Question 1
4 / -1

Two charges q and -3q are fixed on x-axis separated by distance d. Where should a third charge 2q be placed from A such that it will not experience any force?

A

B

C

D

Solution

Let a charge 2q be placed at P, at a distance l from A where charge q is placed, as shown in figure.
The charge 2q will not experience any force, when force of repulsion on it due to q is balanced by force of attraction on it due to -3q at B where AB = d
or
(l + d)² = 3l² or 2l² - 2ld - d² = 0

If charge 2q is placed between A and B then
Question 2
4 / -1

Consider the charges q, q and -q placed at the vertices of an equilateral triangle of each side l. The sum of forces acting on each charge is

A

B

C

D
zero

Solution

From diagram, force on q₁(= q) at A,

where  is the unit vector along BC
Force on
(where   is the unit vector along AC)
Force on q₃(= -q) at C,

where  unit vector along the direction bisecting ∠BCA
Question 3
4 / -1

A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in figure. The ratio of charges Q and q is

A
4

B
1/4

C
-4

D
-1/4

Solution

For the system to be in equilibrium, net force on q = 0 or
or Q = (-q)/4 or Q/q = -(1/4)
Question 4
4 / -1

Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in figure. The force on a charge of 1μC placed at the centre of square is

A
7N

B
8N

C
2N

D
zero

Solution

Forces of repulsion on 1μC charge at O due to 3 μC charge, at A and C are equal and opposite. So they cancel each other.
Similarly, forces of attraction of 1μC charge at O due to -4μC charges at B and D are also equal and opposite. So they also cancel each other.
Hence the net force on the charge of 1μC at O is zero.
Question 5
4 / -1

Three charges of equal magnitude q is placed at the vertices of an equilateral triangle of side l.The force on charge Q placed at the centroid of the triangle is

A

B

C

D
zero

Solution

As shown in figured raw AD ⊥ BC.

Distance AO of the centroid O from A

∴ Force on Q at O due to charge q₁ = q at A,

Similarly, force on O due to charge q₂ = q at B
along BO and force on Q due to charge q₃ = q at c

Angle between forces F₂ and F₃ = 120°
By parallelogram law, resultant of along OA
∴ Total force on Q =