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Grade 12Electrostatics

1.Two point charges placed at a distances of 20 cm in air repel each other with a certain force. When a dielectric slab of thickness 8 cm and dielectric constant K is introduced between these point charges, force of interaction becomes half of it’s previous value. Then K is approximately

Profile image of Akshat Jain
11 Years agoGrade 12
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3 Answers

Profile image of noogler
ApprovedApproved Tutor Answer11 Years ago
= 4
on introducing slab distance btwn charges reduces to (20-8)+8(k)1/2 =rl
 F=1/2F‘
Profile image of Akshat Jain
11 Years ago
can you please tell me the whole solution
Profile image of Sowmya yarnagula
7 Years ago
  • We know F=1/4​pi£0 q1q2/r^2
  • F/2=1/4pi£0 q1q2/(distance bw charges without dielectric+thickness of slab under root of dielectric constant-1)^2
  • Substituting nd solving gives the value 4

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