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Kipish [7]
3 years ago
13

Assume charge builds up on the cloud and on the ground until a uniform electric field of 4.00 ✕ 106 N/C throughout the space bet

ween them makes the air break down and conduct electricity as a lightning bolt. What is the maximum charge the cloud can hold?
Physics
1 answer:
kumpel [21]3 years ago
6 0

Answer:

The value of the maximum charge the cloud can hold  Q = 177.08 coulomb

Explanation:

Given data

Electrical Field E = 4 × 10^{6} \frac{N}{C}

the maximum charge the cloud can hold is given by

Q = E A x ---- (1)

where A = area  = 5 × 10^{6}

x = electrical Permittivity of air = 8.854 × 10^{-12}

Q = (4 × 10^{6} ) (5 × 10^{6} ) ( 8.854 × 10^{-12} )

Q = 177.08 coulomb

This is the value of the maximum charge the cloud can hold.

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Not in book
umka2103 [35]

Answer:

x=2.4365\ m

and

x=-1.4365\ m

Explanation:

Given:

  • first charge, q_1=5\times 10^{-3}\ C
  • second charge, q_2=3\times 10^{-3}\ C
  • position of first charge, x_1=-2\ m
  • position of second charge, x_2=-1\ m

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

E_1=E_2

  • since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}

\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}

3(r^2+1+2r)=5r^2

2r^2-6r-3=0

r=3.4365 \&\ r=-0.4365

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

x=-1+3.4365=2.4365\ m

and

x=-1-0.4365=-1.4365\ m

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