Let's analyze each part step by step: --- ## **First Comparison: Cases A and B** - **Case A:** Two positive charges (+Q) at a distance d from point X. - **Case B:** One positive and one negative charge (+Q, -Q) at the same distance d from point X. **Electric field at X:** - In **Case A**, both electric fields add up since both are in the same direction. - In **Case B**, the fields are in opposite directions and partially cancel out. **Therefore, the electric field at point X in case A is:** - **(A) greater than the electric field at point X in case B.** --- ## **Second Comparison: Cases C and D** - **Case C:** Two positive charges (+2Q), each at distance d from X, at an angle α from the x-axis. - **Case D:** Four positive charges (+Q), each at the same distance and angle. **Electric field at X:** - Each charge in case D is +Q, but there are 4 of them. - In case C, there are 2 charges of +2Q. - Both cases have the same total charge (4Q) and symmetry. **The vector sum of the fields will be the same (since 2 × 2Q = 4 × Q) and the geometric configuration is symmetric.** **Therefore, the electric field at point X in case C is:** - **(C) equal to the electric field at point X in case D.** --- ## **Third Comparison: Cases E and F** - **Case E:** Single charge +Q at distance a from X. - **Case F:** Ten positive charges, each +Q/10, distributed evenly on a circle of radius a centered on X. **Electric field at X:** - In case F, symmetry causes the electric field contributions from each charge to cancel out, resulting in zero net field. - In case E, there is a nonzero field from the single charge. **Therefore, the electric field at point X in case E is:** - **(A) greater than the electric field at point X in case F.** --- ### **Summary of Answers:** 1. **First question:** A 2. **Second question:** C 3. **Third question:** A