A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate syst

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A point charge +q is placed at the origin. A second point charge +9q is placed at (d, 0, 0) in Cartesian coordinate system. The point in between them where the electric field vanishes is:

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Given: A point charge +q is at the origin and another point charge +9q is at (d,0,0). Find: The point between the charges where the net electric field is zero. Use the principle of superposition. Let the required point be P(x,0,0), where 0<x<d. At point P, the field due to +q is along the positive x-axis and the field due to +9q is along the negative x-axis, so for cancellation their magnitudes must be equal. The electric field due to a point charge is E= kqr^2 Therefore, kqx^2= k(9q)(d-x)^2 Cancelling kq, 1x^2= 9(d-x)^2 Taking the positive root, since x and (d-x) are both positive, 1x= 3d-x So, d-x=3x d=4x x= d4 Hence, the point is ( d4,0,0). Therefore, the correct option is B.

Given: q1=+q at (0,0,0) and q2=+9q at (d,0,0). Find: The coordinate of the point where the electric field vanishes. The problem asks for a point on the line segment joining the two charges. Since both charges are positive, the net electric field can vanish only at a point between them, where the two fields are opposite in direction. Let the point be P(x,0,0) with 0<x<d. Field due to the charge at the origin: E1=k qx^2 This field is directed along the positive x-axis. Field due to the charge at (d,0,0): E2=k 9q(d-x)^2 This field is directed along the negative x-axis. For net field to be zero, E1=E2 So, k qx^2=k 9q(d-x)^2 Cancelling common factors, 1x^2= 9(d-x)^2 Taking square roots, 1x= 3d-x Because 0<x<d, both denominators are positive, so only the positive sign is physically valid: 1x= 3d-x Cross-multiplying, d-x=3x d=4x x= d4 Thus, the required point is ( d4,0,0). Therefore, the electric field vanishes at ( d4, 0, 0).

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