SATHEE: Electrostatics Question 414

Electrostatics Question 414

Question: Two charges $+ 5 μ C$ and $+ 10 μ C$ are placed 20 cm apart. The net electric field at the mid-Point between the two charges is [KCET (Med.) 2000]

Options:

A) $4.5 \times 10^{6}$ N/C directed towards $+ 5 μ C$

B) $4.5 \times 10^{6}$ N/C directed towards $+ 10 μ C$

C) $13.5 \times 10^{6}$ N/C directed towards $+ 5 μ C$

D) $13.5 \times 10^{6}$ N/C directed towards $+ 10 μ C$

Show Answer

Answer:

Correct Answer: A

Solution:

$E_{A}$ = Electric field at mid-point M due to + 5mC charge

$= 9 \times 10^{9} \times \frac{5 \times 10^{- 6}}{{(0.1)}^{2}} = 45 \times 10^{5} N / C$

$E_{B}$ = Electric field at M due to +10mC charge

$= 9 \times 10^{9} \times \frac{10 \times 10^{- 6}}{{(0.1)}^{2}} = 90 \times 10^{5} N / C$

Net electric field at $M = | {\vec{E}}_{B} | - | {\vec{E}}_{_{A}} |$

$= 45 \times 10^{5} N / C = 4.5 \times 10^{6} N / C,$

in the direction of EB i.e. towards + 5mC charge

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Electrostatics Question 414

Question: Two charges +5μC and +10μC are placed 20 cm apart. The net electric field at the mid-Point between the two charges is [KCET (Med.) 2000]

Options:

Answer:

Solution:

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Question: Two charges $+ 5 μ C$ and $+ 10 μ C$ are placed 20 cm apart. The net electric field at the mid-Point between the two charges is [KCET (Med.) 2000]