Continuous Charge Distribution

Correct answer: C. Local density is zero at every point

Explanation: The governing idea is that average linear charge density represents a signed average, not the value at every location. A zero average can result when positive and negative contributions cancel, so the net charge may be zero and both signs may occur. However, the local density can vary from point to point and need not vanish everywhere. Therefore option C is the conclusion that is not always valid; zero average must not be confused with uniform zero density.

Correct answer: B. Two coulomb per metre

Explanation: Step 1: New charge is two hundred eighty coulomb and new length is one hundred ninety six metre. Step 2: Linear density is charge divided by length. Step 3: The exact value is ten by seven coulomb per metre so among the given options two is the closest.

Correct answer: A. Toward the removed part

Explanation: The governing idea is superposition and cancellation by symmetry. For the complete uniformly charged ring, the field of the small removed element was exactly cancelled by the field of the rest of the ring, so the net field was zero. After removal, the remaining field equals the negative of the removed element’s field. At the centre, the field due to a positive element points away from that element; therefore the negative of this field points toward the removed part. Hence option A is correct; option B reverses the vector reasoning, while C ignores the broken symmetry.

Correct answer: B. The opposite direction

Explanation: The result follows from superposition and the reversal of electric-field direction for opposite charge signs. In a complete negative ring, fields cancel at the centre. Removing one negative element leaves the negative of that element’s original field as the residual field. If the same element were positive, its field would reverse direction, and the residual field would reverse as well. Thus the negative-ring case has the opposite direction to the positive-ring case. Option B is correct; A ignores charge sign, C ignores the missing part, and D is unnecessary because the geometry is specified.

Correct answer: A. Because the complete opposite half is absent

Explanation: The governing concept is vector superposition together with geometric symmetry. In a complete uniformly charged ring, every small charge element has a diametrically opposite partner whose field at the centre cancels it. A semicircular arc contains only one half of that pairing structure, so the opposite elements needed for complete cancellation are missing. The individual element fields are not zero, and the arc carries nonzero charge. Therefore the net field is generally nonzero, making option A correct; B, C, and D contradict the physical setup.

Correct answer: A. Because transverse components cancel by symmetry

Explanation: The governing concept is cylindrical symmetry and vector addition. Consider two equal charge elements located symmetrically on opposite sides of the ring. At an axial point, they are equally distant, so their field magnitudes are equal. Their components perpendicular to the axis point in opposite directions and cancel pairwise. Their axial components point in the same direction and add. Repeating this pairing around the ring leaves only the axial component. Thus option A is correct; the other choices confuse charge, vector nature, or geometry.

Correct answer: A. Equal small charge elements occur at equal distances on both sides

Explanation: The key principle is mirror symmetry of the continuous linear charge distribution. For every small element on one side of the rod’s midpoint, there is an equal element at the same distance on the other side. Their field components parallel to the rod are equal and opposite, so those components cancel. Their components perpendicular to the rod point in the same direction and add. Consequently, the net field lies along the perpendicular bisector. Option A states the required condition; the other options are irrelevant or physically false.

Correct answer: C. They become unequal and may leave a residual component

Explanation: Complete cancellation of the components parallel to the rod requires equal charge contributions from corresponding elements on both sides. Increasing the right-half density breaks that mirror symmetry. The corresponding elements then produce unequal fields, so their components along the rod no longer cancel exactly. A residual component can therefore remain, with its direction depending on which side has the greater effective contribution and on the sign of charge. Option C is correct; A assumes preserved symmetry, B overstates cancellation, and D confuses component direction with distribution type.

Correct answer: C. Add the charges of small area elements

Explanation: The governing relation for a surface distribution is dq = sigma dA, where the surface charge density sigma may vary from point to point. Since density decreases with radial distance, one constant value cannot correctly represent the entire surface. The surface must be divided into small area elements, their individual charges sigma(r)dA evaluated, and the contributions integrated: Q = integral sigma(r)dA. Therefore option C is correct. Options A and B use only one local value, while D has no physical basis.

Correct answer: A. Spherical symmetry

Explanation: Symmetry is identified by the variables on which the charge density depends. If rho is a function only of the radial distance r, then all points on a sphere of radius r have the same density, regardless of direction. Rotating the distribution about the centre does not change it, which is precisely spherical symmetry. Option A is therefore correct. Linear or square symmetry does not describe a three-dimensional radial distribution, and option D contradicts the stated directional independence.

Correct answer: A. It increases more rapidly

Explanation: Linear charge density is charge per unit length, so the charge in a small equal-length section is approximately dq = lambda(x) dx. If lambda(x) is proportional to x squared, then sections located at larger x have larger density and therefore larger charge for the same dx. Moreover, the rate of increase itself grows with distance because a quadratic dependence is faster than a linear one. Hence option A is correct. B would require constant density, while C and D contradict the stated positive quadratic growth.

Correct answer: A. Because positive and negative charges act from different positions

Explanation: The governing distinction is between net charge and the spatial distribution of charge. Zero total charge means only that the algebraic sum of positive and negative charges is zero. In a dipole-like arrangement, the charges occupy different positions, so at a general observation point their distances and field directions are different. Their electric-field vectors therefore do not cancel everywhere, although they may cancel at selected points because of symmetry. Option A is correct; B contradicts the premise, and C and D misstate charge density or field nature.

Correct answer: A. Identify the distribution type, then density, symmetry, and vector direction

Explanation: The governing method for a continuous charge distribution is to model the charge correctly before integrating its electric-field contributions. First identify whether the charge lies on a line, surface, or volume and choose the corresponding density. Next inspect symmetry to simplify the vector sum. Finally retain the direction of each contribution while integrating. Option B loses spatial information, C incorrectly discards vectors, and D can produce dimensional errors.

Correct answer: A. Because contributions of infinitesimal elements are added in the limiting process

Explanation: Integration is called a limiting sum because a continuous distribution is conceptually divided into many small charge elements. For example, a line element carries dq = lambda dx, and the field is obtained by summing dE over all elements. As the element size approaches zero and the number of elements approaches infinity, the discrete sum becomes the integral. Option B is not required, while C and D misunderstand vector addition and density.