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BigorU [14]
3 years ago
13

Compare the signs of ƒ for lenses and mirrors.

Physics
2 answers:
ELEN [110]3 years ago
4 0

Answer:

A positive focal point for concave mirrors and convex lenses.

A negative focal point for convex mirrors and concave lenses.

Explanation:

STALIN [3.7K]3 years ago
3 0

Answer:

simple

Explanation:

<h3>CONCAVE MIRRORS AND LENSES</h3>

<h3>f= negative</h3>

<h3>CONVEX MIRRORS AND LENSES</h3><h3 /><h3>f= positive</h3>

<h3>PLEASE FOLLOW ME AND MARK IT BRAINLIEST</h3>

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
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Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

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Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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