Question

A small sphere of charge +6x10-6 C is suspended by a string of negligible mass. A charge of -9.0x10-6 C is placed directly to the right of the sphere and 0.22 m away from it. The string is deflected 5° from the vertical. Find the tension in the string.

Answer 1

Answer:

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Answer 2

Answer:

The tension in the string will be 2×10¹²N.

Explanation:

Since the string is making some angles with the vertical we will have two components of tension. One in the vertical direction and the other one in the horizontal direction.

For vertical direction

The tension in the vertical direction will be balanced by the weight of the sphere.i.e.

Tcosθ=mg     (1)

But is mentioned in the question that the to consider mass as negligible. So, the tension in the vertical direction will become zero.

For horizontal direction

The tension in the horizontal direction will be balanced by the force acting between the sphere and the charge -9.0×10⁻⁶C .i.e.

Tsinθ=Fe     (2)

where,

$Fe=\frac{k\times q1q2}{r^2}$    (3)

q₁= charge on the sphere

q₂=the other charge placed

r= distance between the charges

k=9 × 10⁹Nm²/C²

Values given in the question are,

q₁= 6×10⁶C

q₂=-9.0×10⁻⁶C

r=0.22 m

θ=5°

For small  angled θ,sinθ≈θ

By using equations (2), (3), and values that are given in question we get;

$T\times 5=\frac{9 \times 10^9\times6\times10^6\times-9.0\times10^-6}{(0.22)^2}$$=1\times 10^1^3$

$T=\frac{1\times 10^1^3}{5}=2\times 10^1^2N$

Hence, the tension in the string will be 2×10¹²N.