The charge of sphere 2 is twice that of sphere 1. If F is the force of sphere 1

Question

The charge of sphere 2 is twice that of sphere 1. If F is the force of sphere 1 on sphere 2 with a separation of r, what is the force of sphere 2 on sphere 1 if the separation is now reduced to r/2? A. 16F B. 8F C. 4F D. F E. F/4 The electric field at the dot is A. 10 i B. -10 i V/m C. 20 i V/m D. 30 i V/m E. 60 i V/m Draw the locations of the following charges and the electric fields for extra credit (10) Four equal positive point charges are located at the comers of a square, their positions in the xy-plane being (1, 1), (-1, 1), (-1, -1), (1, -1). The electric field on the x-axis at (1, 0) points in the same direction as: A. i. B. i. C. -i D. k. E. -j. What is the minimum magnitude of an electric field that balances the weight of a plastic sphere of mass 10 g that has been charged to -5.0 mu C? A. 6.4 time 10^6 N/C B. 4.5 times 10^6 N/C C. 2.0 times 10^6 N/C D. 2.1 times 10^-7 N/C E. none of these If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region. A. True B. False C. Need more information If the electrical potential in a region is constant, the electric field must be zero everywhere in that region. A. True B. False C. Need more information A positive charge, if free, will tend to move. A. away from infinity. B. from high potential to low potential C. toward infinity. D. in the direction of the electric field. E. from low potential to high potential If the electric potential in a region is given by V(x) = 4/x^3, the x component of the electric field in this region is A. -18x B. 12x C. 12x^-4 D. -12x^-2 E. 18x^-4.

Explanation / Answer

1. Coulomb force is inversely proportional to square of distance. Hence if r is reduced by r/2 then force must be 4 times. so, answer is C

2. From the diagram we observe that potential varies linearly with x, hence

let V(x) = m*x + C ...........(i)

at x=4cm , V = 0 => 4m +C= 0 .........(ii)

at x=2cm, V = 20 volt=> 2m + C = 20 .................(iii)

Solving (ii) and (iii) we get , m = -10 and C = 40

hence V(x) = -10x + 40 .............(iv)

Since E = -dV/dx=> E = -(-10) V/m along +ve x, hence E = 10 i V/m.

Hence answer = A

3. Electric field at P(1,0) due to q1(1,1) and q4(1,-1) will be equal and opposite, hence they will cancel each other.

Y component of E2 and E3 will be cancelled out hence net electric field P must be along positive x-direction. Hence answer is B

4. q*E = m*g => E = m*g/q => E = 10*10-3 kg * 9.8 ms-2 / (5.0*10-6C) => E = 1.96*104 N/C.

Hence answer would be E(Non of these)

5. since E = - dV/dr=> dV/dr = 0 , hence V = constant (may or may not be zero) Hence answer is more info needed

6. Answer is true , because derivative of a constant is zero

7. This question has multiple answers. Positive charge moves along the electric field lines and since electric filed lines are directed from high to low potential hence we can say that positive charge moves from high to low potential. Hence answers are B and D

8. Since E = - dV/dx, hence E = - d(4/x3) /dx => E = 12/x4 , hence answer is C

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